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Chapter 15. Vector Calculus

Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

http://matrix.skku.ac.kr/Cal-Book1/Ch5/

Chapter 15. Vector Calculus

15.1 Vector Differentiation  http://youtu.be/q0aVmUCXgTI

문제풀이 by 김동윤  http://youtu.be/iSUME4Q1WPM

15.2 Line Integrals   http://youtu.be/wHINlpNXYaU

문제풀이 by 김범윤  http://youtu.be/ZdRjCfJeHM8

15.3 Independence of the Path   http://youtu.be/jGGOL3QDj1Y

문제풀이 by 김유경  http://youtu.be/TreCe8ESEiU

15.4 Green’s Theorem in Plane  http://youtu.be/WxdTbaSb_ZI

문제풀이 by 서용태  http://youtu.be/wLTHYaANwtI

15.5 Curl and Divergence  http://youtu.be/IswmJUCTeNA

문제풀이 by 오교혁  http://youtu.be/j7F3xVNdHvA

15.6 Surface and Area   http://youtu.be/xX6tNVpegbs

15.7 Surface Integrals  http://youtu.be/nrzIrM4doLo

문제풀이 by 이원준  http://youtu.be/s_MRgW2By38

15.8 Stokes’ Theorem   http://youtu.be/t4skc_PzJvg

15.9 Divergence Theorem  http://youtu.be/3BmcFr81kuQ

문제풀이 by 최주영  http://youtu.be/vGMLoGWF1Is

15.1 Vector Differentiation

Vector Fields

Associated with every point in a region we can imagine both a direction and a magnitude about gravitational force, the velocity of a flowing fluid. They are expressed by a vector at each point in their region, which is producing a vector field.

DEFINITION1

Let be a subset of (or ). A scalar field on is a scalar valued function (or ) that assigns to each point (or ) in .

Any real valued function on is a scalar field. For example a function given by is a scalar field, defined on , is a scalar field defined on . A scalar field can be called a scalar function.

DEFINITION2

Let be a subset of (or ). A vector field on is a function that assigns to each point (or ) in , a two (or three)-dimensional vector (or ).

The best way to understand a vector field is to draw the arrows for the vector at a few representative points . Since is a two dimensional vector, we may write it as follows:

where and are scalar functions of two variables, and where and are unit vector along the coordinate axes. These scalar functions are called component functions of the vector field. We can also plot the vector field in two or three dimensions with the aid of a computer. Since a computer can plot a large number of vectors, this gives a better impression of the vector field than drawing by hand.

EXAMPLE 1

Draw the vector field on defined by .

Solution. The length of the vector is . Vector points roughly away from the origin and vectors farther from the origin are longer.

(See Figure 1.)

var('x,y')

plot_vector_field((1/2*x, y), (x,-3,3), (y,-3,3))

Figure 1       ■

EXAMPLE 2

Draw the vector field on defined by .

Solution.  The length of the vector is .

Figure 2

EXAMPLE 3

Draw the vector field on defined by .

Solution.  At each point , is a vector of length . For , all vectors are in the direction of the negative -axis, while for , all vectors are in the direction of the positive -axis. In each plane , all the vectors are identical.

var('x, y, z')

plot_vector_field3d((0,0,-y), (x,-1,1), (y,-1,1), (z,-1,1))

Figure 3

DEFINITION3

If is a scalar function of two variables, then the gradient of , denoted by grad (or ), is defined by the vector

Likewise, if is a scalar function of three variables, then the gradient, grad (or ), is defined by

Therefore, is a vector field in (or ) and it is called a gradient vector field.

EXAMPLE 4

If is defined to be , then the grad is a vector field on .

If is defined to be , then the grad is a vector field on .

EXAMPLE 5

Find the gradient vector field of . Draw the grad together with level curves of . How are they related?

Solution. The gradient vector field of is given by the grad .

Figure 4 shows the contour map of with the gradient vector field. Notice that the gradient vectors are perpendicular to the level curves.

http://matrix.skku.ac.kr/cal-lab/Sec15-1-Exm-5.html

var('x, y')

f = (x^2 - y^2)

plot_vector_field(f.gradient(), (x, -2, 2), (y, -2, 2), color='blue')

Figure 4 Contour map and gradient vector field of

If you look at the Figure 4, it is clear that the gradient is perpendicular to the level curve. This can be proved easily.

Let be a function having the first order continuous partial derivatives. Let be the level curve at the level . Then on . Let be a curve in , then . Hence, using the chain rule, we have .  So,  . This means is orthogonal to the tangent for all real value .       ■

15.1 EXERCISES (Vector Differentiation)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-1-Sol.htm

1-7. Sketch the vector field .

1.

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-15-1-1.html

var('x,y')

vf=plot_vector_field((3,4), (x,-3,3), (y,-3,3), aspect_ratio=1);

show(vf)

2.

Solution.

http://matrix.skku.ac.kr/cal-lab/Sec15-1-2.html

var('x,y')

vf=plot_vector_field((1/2*x,-2*y), (x,-3,3), (y,-3,3), aspect_ratio=1);

show(vf))

3.

Solution.

var('x,y')

vf=plot_vector_field((5*y,1/2), (x,-3,3), (y,-3,3), aspect_ratio=1);

show(vf)

4.

Solution.

var('x,y')

vf=plot_vector_field((x+y,x), (x,-3,3), (y,-3,3), aspect_ratio=1);

show(vf)

5.

Solution.

http://matrix.skku.ac.kr/cal-lab/Sec15-1-Exs-5.html

var('x,y')

vf=plot_vector_field((y/sqrt(x^2+y^2),-x/sqrt(x^2+y^2)), (x,-3,3), (y,-3,3),

aspect_ratio=1);

show(vf)

6.

Solution.

http://matrix.skku.ac.kr/cal-lab/Sec15-1-6-7.html

var('x,y')

vf=plot_vector_field((y/sqrt(x^2+y^2),x/sqrt(x^2+y^2)), (x,-3,3),

(y,-3,3), aspect_ratio=1);

show(vf)

7.

Solution.

http://matrix.skku.ac.kr/cal-lab/Sec15-1-6-7.html

var('x,y,z')

plot_vector_field3d((0,0,1), (x, -3,3), (y,-3,3), (z,-3,3))

15.2 Line Integrals

In this section, the concepts of (ordinary) integral calculus are extended to vector functions. Line integrals are useful in the calculation of work done by variable forces along paths in space.

For the ordinary definite integral the region of integration is an interval on the -axis. That is, we integrate along the -axis from to . This concept can be generalized to define a definite integral evaluated along a curve.

Line Integral of Scalar Fields

Let be a continuous function defined on a domain . Let be a smooth curve in defined as . We wish to define the integral of along the curve .

For this, we divide the integral into equal parts . The corresponding points on the curve are , in Figure 1.

Figure 1

Note that divide the curve into sub-arcs. We assume that the length of these sub-arcs are , respectively.

In each sub-arc, choose a random point which corresponds to a point in the subinterval of .

If we draw a rectangle with the base of and height , then the area of this rectangle is in Figure 2.

Figure 2

Now we form at the sum

.

If exists, then we say that the line integral of with respect to along the curve from to exists. In this case, this limit is denoted by

or   .

This gives us the following definition.

DEFINITION1

Let be a continuous function defined on a domain . Let be a smooth curve. Then the line integral of along is defined as

given that the limit exists.

The line integral is the surface area of the surface above the curve up to under the surface .

We shall assume that line integrals are evaluated along a curve which has  an arc length parameterization. In the case of a parameterization with respect to arc length in Section 12.3, we have

.

Therefore, we have the following theorem which gives a working rule to evaluate a line integral along a curve.

THEOREM 2

Let   be a continuous function defined on a domain and be a smooth curve with arc length parameterization in . Then

.

Let us look at some examples.

EXAMPLE 1

Evaluate where is the semi-circle .

Solution.

■

EXAMPLE 2

Evaluate where is a circle .

Solution.

■

A curve is said to be a closed curve (path) when the endpoints coincide. That is, is a closed curve if . If is a closed curve, then the line integral is denoted by .

Properties of Line Integrals of Vector Fields

Let be a vector field. Then a line integral of along (taken over) the curve is defined as

.

Observe that evaluation of a line integral reduces to evaluation of an ordinary integral.

For the line integral , the following properties follow from integral calculus:

1. where is a constant,

2. ,

3. ,

where is the sum of two curves and .

4. , where is a curve traversed in

opposite direction of . (See Figure 3.)

Figure 3

EXAMPLE 3

Find the line integral of along

(a) the horizontal line .

(b) the vertical line . (See Figure 4.)

Figure 4

Solution. (a)

.

(b)

.

Moreover we consider that a smooth curve is given by the vector equation

, .

If is the function of three variables that is continuous on a region including , then we define the line integral of

.

Next we have a simple example.

EXAMPLE 4

Evaluate , where is a circular helix given by the vector equation , . (See Figure 5.)

Figure 5

Solution.

.     ■

15.2 EXERCISES (Line Integrals)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-2-Sol.htm

1-3. Evaluate the following line integrals:

1. Evaluate , where is the upper half of the unit circle .

t=var('t')

half_circle=parametric_plot((cos(t),sin(t)),(t,0,pi()));

show(half_circle, aspect_ratio=1)

integral(1+cos(t)^3*sin(t)^2*sqrt(diff(cos(t),t)^2+diff(sin(t),t)^2),t,0,pi())

2. , where is the arc of the circular helix from to .

Solution. Let

.

3. Let . Find where is any smooth curve from to .

Solution.

4. If , evaluate .

var('t')

A=vector([t,-t^2,t-1])

B=vector([2*t^2,0,6*t])

v=A.dot_product(B)

v.integrate(t,0,2)

5. If , , then evaluate

a. ,

b. .

6. If , then evaluate .

7-9. (Line Integrals) Given the following vector fields and oriented curves , evaluate .

7. on the semicircle for .

8. on the line segment from to .

9. on the line

15.3 Independence of the Path

Potential Functions

We know that if is a scalar function then the grad is a vector field. Therefore, given a vector field , we can construct the scalar function with the property that

.

If such a scalar function exists, then is said to be conservative.

DEFINITION1

Let be a vector field. Then is called conservative if there is a scalar function such that

.

A function is called a potential function (or scalar potential) for .

Note that if a scalar function is a potential function of a vector field , then is also a potential function of for any constant . Not all the vector fields are conservative, but such fields arise frequently in physics. The most prominent examples of conservative forces are the force of gravity and the electric field associated to a static charge.

According to Newton's Law of Gravitation, the gravitational force field between two objects with masses m and M is given by

where is the position vector of the object with mass from the other object with mass and where is the gravitational constant. It is easy to check that

where .

Therefore, the gravitational force field is conservative and is a potential function for .

If is conservative, then

.

That is, and .

Now take a partial derivative with respect to of , and with respect to of to get    and

In addition, if we assume that the second order partial derivatives are the same, then we have

This turns out to be both necessary and sufficient condition for a vector field to be conservative, in special type of domain, which are simply connected.

A domain is said to be simply connected if it does not have a hole inside. (However, this is not a rigorous definition). For example, any disk: , upper half plane: are simply connected domains in  , whereas annulus:　 is not simply connected.　(See Figure 1.)

Simply connected

Not simply connected

Figure 1

THEOREM 2

If and have continuous first order partial derivatives and the vector field is conservative, then, in simply connected domains, we have

.

The converse of Theorem 2 is in general not true. It is true only for special domains that we will see later. (See Example 2.) In simply connected regions, we can state the converse of Theorem 2 for verifying that a vector field on is conservative. We will check the proof in a result of Green Theorem in Section 15.4.

THEOREM 3

Let be a vector field on an open simply connected region . Suppose that and have continuous first order partial derivatives and . Then isconservative.

Suppose that is a vector field on a simply connected domain where , and have continuous first order partial derivatives. Then is conservative if and only if , and . The proof is similar to Theorem 2 and 3.

EXAMPLE 1

Determine which of two vector fields are conservative

(1)

(2)

(3) .

Solution.  For (1) we find , .

Since they are not equal, the vector field is not conservative.

var('x,y')

F=vector([4*x*y,-x^3])

print bool(diff(F[1],x)==diff(F[0],y))

For (2) we find , .

Since they are equal, the vector field is conservative.

For (3) we find , .

Since they are not equal, is not conservative.

var('x,y,z')

H=vector([2*x,-z,sin(z)])

print bool(diff(H[2],y)==diff(H[1],z))

print bool(diff(H[2],x)==diff(H[0],z))

print bool(diff(H[1],x)==diff(H[0],y))

Reference:

Independence of Path

Let and be two points in an open region . For some special fields, the line integral of the vector field has the same value for all paths from to . (See Figure 2.) Then the line integral is said to be the independence of path in .

Figure 2  Independence of Path

There is an useful property of line integral in a conservative field when the path of integration is a closed curve: the vector field is conservative on the domain if and only if around every closed curve in the domain .

If is a conservative vector field with potential function , then the line integral from to is independent of path joining to . (See Figure 2.)

.

Thus the line integral depends only on the end points and and not on the path joining them. Recall that when , then . In such a case, is called a conservative vector field and is called its scalar potential (potential function).

Note that a conservative force field is also irrotational (since ).

EXAMPLE 2

The vector field on except the origin has no potential function, although ,

and .

Suppose that a closed curve is the unit circle : , . We evaluate around . First we have

.

Then .

Since  , the vector field is not conservative.

For a conservative vector field, how do we find the potential function? Let us look at some examples.

EXAMPLE 3

Test the existence of a potential function for the vector field on its domain.

Solution. Since and , that is, , the vector field is not conservative. So, there is no potential function.   ■

EXAMPLE 4

What is the potential function for the conservative vector field

on ?

Solution.  From Theorem 3, we know that is conservative. Suppose for some scalar function . Then we have

.

Integrating both sides with respect to we get

,

where is a constant of integration and a function of .

Now differentiate with respect to , we get

.

Thus  .

Integrating with respect to , we get

( is a constant).

Since we just want “a” potential function not “the” general potential function, we do not need a constant of integration here. Putting it all together, we get the potential function

.

x = var('x')

y = function("y",x)

M = 1+3*x^2*y

N = x^3-2

solution=desolve(diff(y,x)==-M/N,y)

f=simplify(solution==y)*solution.denominator()

potential=f.lhs()-f.rhs()

potential    # This is a potential function for F

Answer :  (x^3 - 2)*y - c + x

Work Done by a Force (Work Integral)

THEOREM 4  Path Independence Theorem

Let be a vector field such that and let and be two points of . Let be a potential function for . For any piecewise smooth curve from and ,

Thus the line integral in this case depends only on the end points and and not on the curve . Thus we may write

A natural application of the line integral is to define the work done by a force in moving (displacing) a particle along a curve from point to point as

work done .

When denotes the velocity of a fluid, then the circulation of around a closed curve is defined by .

1. The work done by a conservative force field in moving a particle from to is independent of the path joining and , that is, it depends only on the end points and . In such cases a scalar potential exists that is, . Thus the work done from to equals .

2. If is a closed smooth curve and is a conservative field, then

along any closed curve because

which follows from the independence of the path. (See Figure 3.)

Figure 3

EXAMPLE 5

If , evaluate the line integral around a triangle in the -plane with , , . (See Figure 4.)

Figure 4

(a) in the counterclockwise direction (b) what is the value in the opposite direction?

Solution.

(a) In the counterclockwise direction:

.

Along : , varies from 0 to 2, since ,

and ,

.

Along : , varies from 0 to 1, since , and

,

.

Along : , varies from 2 to 0, since ,

and ,

.

Thus .

The required line integral is in the counterclockwise direction.

There is an another solution using line segment from a point to other point .

On : , , since and ,

.

On : , , since and ,

.

On : , , since and

,

.

The result is same.

(b) The value of the line integral in the opposite direction is . ■

http://matrix.skku.ac.kr/cal-lab/Sec15-2-Exm-5.html

var('x,y,t')

r=vector([0,0])+t*(vector([2,0])-vector([0,0]))

dr=diff(r,t)

F=vector([2*x+y^2,3*y-4*x]).subs(x=r[0],y=r[1])

integrand=F.dot_product(dr)

line_integral_1=integral(integrand,t,0,1)

line_integral_1

EXAMPLE 6

Find the line integral . where is the rectangular curve from to to . (See Figure 5.)

Figure 5

Solution. Let . Then .

Here is a piecewise smooth curve made up of a horizontal piece

and a vertical piece

.

Along : , varies from 2 to 4, since , and ,

.

Along : , varies from 5 to 6, since , and ,

.

Thus the line integral is

.       ■

EXAMPLE 7

Find the line integral where is a vector field in example 4 and is the path from to .

Solution. Since the vector field is conservative, there is the potential function in example 4. By the Path Independence Theorem,

.   ■

15.3 EXERCISES (Potential Function and Independence of Path)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-3-Sol.htm

1-6. Determine whether or not the vector field is conservative. If it is conservative, find a potential of .

1.

Solution.

http://matrix.skku.ac.kr/cal-lab/Sec15-2-Exs-1.html

var('x,y')

F=vector([y*exp(-x),exp(-x)])

print bool(diff(F[1],x)==diff(F[0],y))  \

Then is not conservative.

2.

Solution.

var('x,y')

F=vector([2*x*y,x^2+2])

print bool(diff(F[1],x)==diff(F[0],y))

x = var('x')

y = function("y",x)

M=2*x*y

N=x^2+2

solution=desolve(diff(y,x)==-M/N,y)

f=simplify(solution==y)*solution.denominator()

potential=f.lhs()-f.rhs()

potential

Answer :  -(x^2 + 2)*y(x) + c

3.

Solution. Let , and .

Since and ,

Then is not conservative.

var('x,y,z')

F=vector([2*y^2*z^3,2*x*y*z^3,3*x*y^2*z^2])

print bool(diff(F[2],y)==diff(F[1],z))

print bool(diff(F[2],x)==diff(F[0],z))

print bool(diff(F[1],x)==diff(F[0],y))

4.

Solution. Since , and , is conservative.

To find the potential function , first we begin

―(1)

Taking partial derivative of with respect to , we get  . Hence from (1) we get       ―(2)

Then   ―(3)

Now taking partial derivative of with respect to we get . Hence from (3) we get     , a constant. Therefore, .

var('x,y,z')

F=vector([exp(z),1,x*exp(z)])

print bool(diff(F[2],y)==diff(F[1],z))

print bool(diff(F[2],x)==diff(F[0],z))

print bool(diff(F[1],x)==diff(F[0],y))

var('x,y,z,c')

fx=exp(z);fy=1;fz=x*exp(z)

f_1=integral(fx,x)    # +g(y,z)

f_1y=diff(f_1,y)     # fy= diff(f_1,y)+gy(y,z)

gy(y,z)=fy-diff(f_1,y)

g(y,z)=integral(gy(y,z),y) # +h(z)

f_2=f_1+g(y,z) #+h(z)

f_2z=diff(f_2,z)     #fz=diff(f_2,z)+ hz(z)

hz(z)=fz-diff(f_2,z)

h(z)=integral(hz(z),z)

f=f_2+h(z)+c

f

Answer :  x*e^z + c + y

5.

Solution. Not conservative.

6.

Solution. Conservative, .

7. Show that the vector field is not conservative.

Solution.

A=vector([exp(x*z),3*x*y*z,2*y])

print bool(diff(A[2],y)==diff(A[1],z))

print bool(diff(A[2],x)==diff(A[0],z))

print bool(diff(A[1],x)==diff(A[0],y))

Then is not conservative.

8. Determine whether the force field is a conservative field.

Solution.

var('x,y,z')

F=vector([2*x*z,x^2-y,2*z-x^2])

print bool(diff(F[2],y)==diff(F[1],z))

print bool(diff(F[2],x)==diff(F[0],z))

print bool(diff(F[1],x)==diff(F[0],y))

Then is not conservative field.

9. a. Prove that is a conservative field.

b. Find its scalar potential .

c. Also find the work done in moving an object in this field from to .

Solution. Scalar potential .

Work done.

var('x,y,z')

F=vector([4*x*y-3*x^2*z^2,2*x^2,-2*x^3*z])

print bool(diff(F[2],y)==diff(F[1],z))

print bool(diff(F[2],x)==diff(F[0],z))

print bool(diff(F[1],x)==diff(F[0],y))

var('x,y,z,c')

fx=4*x*y-3*x^2*z^2;fy=2*x^2;fz=-2*x^3*z

f_1=integral(fx,x)    # +g(y,z)

f_1y=diff(f_1,y)      # fy= diff(f_1,y)+gy(y,z)

gy(y,z)=fy-diff(f_1,y)

g(y,z)=integral(gy(y,z),y) # +h(z)

f_2=f_1+g(y,z) #+h(z)

f_2z=diff(f_2,z)     #fz=diff(f_2,z)+ hz(z)

hz(z)=fz-diff(f_2,z)

h(z)=integral(hz(z),z)

f=f_2+h(z)+c

f

Answer : -x^3*z^2 + 2*x^2*y + c

var('x,y,z,c')

f(x,y,z)=-x^3*z^2 + 2*x^2*y + c

workdone=f(1,1,1)-f(0,0,0)

workdone

10. Find the total work done in moving a particle by a force field along the curve   form to .

Solution.

var('t,x,y,z')

x(t)=t^2+1;y(t)=2*t^2;z(t)=t^3

r(t)=(x(t),y(t),z(t))

dr(t)=(diff(x(t),t),diff(y(t),t),diff(z(t),t))

A(t)=(3*x(t)*y(t),-5*z(t),10*x(t))

F(t)=A(t).dot_product(dr(t))

integral(F(t),t,1,2)

15.4 Green’s Theorem in Plane

In this section, we introduce a very important theorem in vector calculus, which is the Green’s Theorem. Green’s Theorem relates a line integral around a simple closed curve and a double integral with the plane region bounded by that curve, (see Figure 1.) that is possible to transform line integrals to double integrals and vice versa.

Here we assume that the curve has positive orientation (travels to counterclockwise on ). That is, when we travel along , the region always lies on the left side of .

Figure 1

Green’s Theorem has several applications such as finding area of a plane regions and work done by a force field.

THEOREM 1  Greens Theorem

If is a closed region in the -plane bounded by a simple closed curve and if and are continuous functions of and having continuous first order partial derivatives in , then

where is traversed in the positive direction. (See Figure 2.)

* George Green (1793-1841) English mathematician.

Figure 2

1. Vector form of Green’s Theorem

Let and . Then .

Thus Green’s Theorem reduces to

.

2. Area of a plane region using Green’s Theorem. Let be a plane region bounded by the simple closed curve . Let be the area of . Let , so that

.

Thus the area of is

.

Figure 3

3. Green’s Theorem can be applied to regions that are not simply connected. For instance, the boundary of the region in Figure 3 consist of two simply closed curves and and assume that the region is always on the left as and are travesed. Then, the positive direction is counterclockwise for but clockwise for . If we divide into two regions and in Figure 3, then, by Green's Theorem,

where and are the boundary curves of and , respectively.

4. Let be a vector field on an open simply connected region and and have continuous first order partial derivatives. If then by Green’s Theorem .

EXAMPLE 1

Verify Green’s Theorem for , where is a positively oriented curve which is the boundary of the triangle with vertices, , and . (See Figure 4 and 5.)

Figure 4                                   Figure 5

Solution. First, we evaluate the given integral as a line integral. Notice that is a union of three curves : , : and : .

Then .

Let us assume that right hand integral is denoted by and left hand integral is denoted by .

Along , since and ,

.

Along , since and , and . Then

.

Along , since and varies from 1 to 0, . Then

.

Hence .

Next, let and , then we get the double integral

.

Hence Green’s Theorem verified by .                  ■

http://matrix.skku.ac.kr/cal-lab/Sec15-4-Exm-1.html

var('x,y')

M=x^2*y;N=x*y^2

curlF=diff(N,x)-diff(M,y)

RIx=integral(curlF,y,0,x)

integral(RIx,x,0,1)

EXAMPLE 2

Verify Green’s Theorem in the plane for where is the boundary of the region defined by : and . (See Figure 6.)

Figure 6

Solution. (a) The left hand side of the Green’s Theorem is the line integral

.

Here consists of the curves , , , so

Along , since , .

.

Along , since and , and . Then

.

Along , since and varies from 2 to 0,. Then

,

Thus .

(b) Here  , , .

So the right hand side of the Green’s Theorem is the double integral given by

.

The region is described with varying from of the lower branch of the parabola to its upper branch while varies from 0 to 2. Thus

Since , the Green’s Theorem is verified.    ■

http://matrix.skku.ac.kr/cal-lab/Sec15-4-Exm-2.html

var('x,y')

assume(x>0)

M=x^2-2*x*y;N=x^2*y+3

curlF=diff(N,x)-diff(M,y)

RIx=integral(curlF,y,-sqrt(8*x),sqrt(8*x))

integral(RIx,x,0,2)

EXAMPLE 3

Evaluate the line integral using Green’s Theorem where is the boundary of the region defined by the curve , and the line . (See Figure 7.)

Figure 7

Solution.   A region bounded by is

Since and , using Green’s Theorem,

.   ■

http://matrix.skku.ac.kr/cal-lab/Sec15-4-Exm-3.html

var('x,y')

assume(x>0)

M=y/(x+1);N=2*x*y

curlF=diff(N,x)-diff(M,y)

RIx=integral(curlF,y,x^2,x)

integral(RIx,x,0,1)

EXAMPLE 4

Use Green’s Theorem to evaluate the line integral

,

where is a region bounded by        and

Solution.

Figure 8

Area of a Plane Region

THEOREM 2

If has a piecewise smooth boundary with positive orientation, then the area of is

.

Proof.  By Green's Theorem,

.

Adding the two, we get the last term in the theorem.

EXAMPLE 5

Use Green’s Theorem to find the area of the ellipse . (See Figure 9.)

Figure 9. Area=

Solution.  The boundary of the ellipse is the parametric curve

.

Note that and , We have

By Theorem 2

.   ■

Let and be polar coordinates defined by . Then by the definition of the total differential.

It is easy to check that . Hence, by Theorem 2, we get

.

For example, we consider the cardioid for for using , we find

.

EXAMPLE 6

Using Green’s Theorem, find the area of the region in the first quadrant bounded by the curves , and . (See Figure 10.)

Figure 10

Solution. By Green’s Theorem the area of the region bounded by a closed curve is given by .

Here, consists of the curves , , . So

.

Along , , varies from to .

.

Along , , varies from to .

.

Along , , varies from to .

.

Then .      ■

15.4 EXERCISES (Green’s Theorem in the Plane)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-4-Sol.htm

1-4. Use Green’s Theorem to evaluate the line integral , when is

1. , where is a circle of , with positive orientation.

Solution.

x,y,r,t=var('x,y,r,t');

p = implicit_plot(x^2+y^2==9, (x,-3,3), (y, -3,3),cmap=["red"]);

p1 = implicit_plot(x==0, (x,-4,4), (y, -4,4));

p2 = implicit_plot(y==0, (x,-4,4), (y, -4,4));

show(p+p1+p2, aspect_ratio=1)

M=-y;

N=x;

integral(integral((diff(N,x)-diff(M,y))*r, r, 0, 3), t, 0, 2*pi)

2. , where is the positively oriented curve which is the boundary of the region bounded by , .

Solution.

.

3. , where is a circle of , with positive orientation.

Solution.

.

4. where is a boundary of the rectangle , , with positive orientation.

Solution.

.

5-8. Verify Green’s Theorem by evaluating the line integral (a) directly, (b) using Green's Theorem, where is:

5. where is a closed curve of the region bounded by and , with positive orientation.

Solution.

.

6. where is a boundary of the region defined by and , with positive orientation.

Solution.

.

7. where is a boundary of the region enclosed by the circles and , with positive orientation.

Solution.

.

8. , where is a circle of , with positive orientation.

Solution.

.

9. where is a square of with positive orientation.

Solution. Let . To evaluate the line integral, we use Green’s Theorem as follows

10. where is the arc of the parabola in the plane from to .

Solution.

var('x,y,z,t')

p_1=implicit_plot3d(y==x^2, (x,0,1), (y, 0,1), (z, 0,4),color="red", opacity=0.4);

p_2=implicit_plot3d(z==2, (x,0,1), (y, 0,1), (z, 0,4),color="yellow", opacity=0.6);

p_3=parametric_plot3d((t, t^2, 2), (t, 0, 1), thickness=5)

show(p_1+p_2+p_3)

Using Green’s Theorem

integral(x^2*x^2+(x-2)*2,x,0,1)

11. Calculate the area bounded by the ellipse . Hence, deduce the area bounded by the circle .

Solution. (a) .

,

Put , area of circle : .

12. Find the area under arc of the cycloid

, .

13. Find the area under of one arch of the asteroid , .

14. Find the area of the loop of the folium of

Descartes , .

Solution.

@interact

def _(a=(1,(-10,20))):

var('t,b')

def GreenThm(F):

assert(len(F) == 2)

return (diff(F[1],t)-diff(F[0],t))

p_1=parametric_plot(((3*a*t)/(1+t^3),(3*a*t^2)/(1+t^3)), (t, 0,2*pi()), color="green");

show(p_1)

x=(3*a*t)/(1+t^3)

y=(3*b*t^2)/(1+t^3)

Anti=GreenThm((x,y)) #xdy-ydx

15. Let be smooth functions satisfying the following differential equations:

, .

Evaluate the line integral , where is traversed in the positive direction of the curve .

Solution. Using Green’s Theorem, we have

.

16. Let be the solution of the system of differential equations

where are smooth functions in variables .

Assume further that is periodic with the periodicity , namely for all . Suppose that the solution curve (phase trajectory) is a simple closed curve in the phase plane (-plane) and is the region bounded by . Prove that

.

Solution.  Recalling Green’s Theorem,

.

Now parametrizing by , the line integral becomes

.

Therefore, we obtain.

17. Evaluate the integral , where is the boundary of the triangular region with corners , , with positive orientation.

Solution.

18. Let be straight lines from to , from to , from to , and from to . Evaluate .

Solution. Let . Then

19. Let and is an any closed curve containing the origin. Find .

Solution. Let’s consider the region which is the annulus as shown in the figure. Let and denote the outer and the inner circles of .

Then by Green’s Theorem, we have

.

Using , we can compute

Thus the vector field is not conservative. Note that curl. Here is not a simply connected domain.

15.5 Curl and Divergence

In this section, we look at two operations, namely curl and divergence, on vector fields. These operations have many applications in physics and engineering.

Curl of Vector Fields

If is a differentiable vector field with , then

curl   where ;

curl

EXAMPLE 1

Find the curl of the vector field .

Solution.  Using , we have

curl

.

THEOREM 1

Let and be differentiable vector fields with and , . Suppose has continuous partial derivatives. Then

(ⅰ) curl = curl + curl

(ⅱ) curl = + .

Proof.  By , we have

(i)

curl + curl .

(ii)

curl .

THEOREM 2

Let be a three dimensional differentiable vector field with continuous second-order partial derivatives. If is conservative, then curl.

Proof.  Since is conservative vector field, there is a differentiable function of three variables satisfying . But we have

curl

.

Hence, curl.       ■

THEOREM 3

Let be a three dimensional differentiable vector field whose components have continuous partial derivatives. If curl, then is conservative.

EXAMPLE 2

Show that the vector field is conservative.

Solution.  Using , we have

curl

.

Thus is a conservative vector field.

var('x,y,z,i,j,k')

M=x; N=y; P=z

curlF=(diff(P,y)-diff(N,z))*i+(diff(M,z)-diff(P,x))*j +(diff(N,x)-diff(M,y))*k

print curlF

print bool(curlF==0)

Divergence of Vector Fields

If is a differentiable vector field with

then the divergence of is the function of three variables defined by

Notice that the curl of a vector field is a vector field but the divergence of a vector field is a real valued function. If we use the notation then .

THEOREM 4

The following properties are true for differentiable vector fields and and a continuously differentiable scalar function .

(ⅰ) divdiv div ,

(ⅱ) divdiv .

EXAMPLE 3

Find the divergence of the vector field where

and

.

Solution.  We take the three partial derivatives and add them. Then,

div .  div .

Hence div .

var('x,y,z')

M=x-3*y; N=x^2*z^2+cos(z); P=x*y*z^2

divF=diff(M,x)+diff(N,y)+diff(P,z)

divF

EXAMPLE 4

Compute the divergence of the vector field

.

Solution.   div = .         ■

If is a vector field on , then curl is also a vector field in . Therefore, we can compute its divergence. The next theorem shows that the divergence of curl is .

THEOREM 5

If is a vector field on and it , and have continuous second-order partial derivatives, then divcurl ) = 0.

Proof.  Using the definition of divergence and curl, we have

div(curl )

.   ■

EXAMPLE 5

Show that the vector field is not the curl of another vector field.

Solution. Note that div. If were the curl of another vector field, then div would be zero by Theorem 4. This is a contradiction to div. Therefore is not the curl of another vector field.

We can rewrite Green’s Theorem using these new ideas. These rewritten versions in turn are closer to some theorems that we will see later. Suppose we write a two-dimensional vector field in the form where and are functions of and . Then

curl ,

and so . Thus Green’s Theorem says

.

This is called the vector form of Green’s Theorem.

Nature of Curl and Divergence of Vector Fields

The concepts about curl and divergence of a vector field are associated with the tendency of the vector field to rotate or to diverge where is the velocity of fluid (or gas) in physics applications. We can understand the terms in the following discussion, by calculating curl and divergence of vector fields. If at a point , then is called irrotational at . If div, then is said to be incompressible.

For the vector field , it is easy to show that . Thus the vector field is irrotational. (See Figure 1.) On the other hand, for the vector field , since , a vector field is said to be imcompressible. (See Figure 2.)

irrotational                           incompressible

Figure 1                       Figure 2

Next we have another differential operator. If is a function of three variables, we have

and  by computing divergence of the gradient of . The operator is called the Laplace operator because of its relation to Laplace’s equation

.

We can also apply the Laplace operator to the vector field in terms of its components:

.

EXAMPLE 6

Find .

Solution.

var('x,y,z,i,j,k')

M(x,y,z)=x*exp(y*z); N(x,y,z)=x*y*z; P(x,y,z)=-3*y^2-z^2

divF=diff(M,x)+diff(N,y)+diff(P,z)

divF

Answer :  x*z - 2*z + e^(y*z)      ■

EXAMPLE 7

Show that the vector field is not conservative.

Solution.

var('x,y,z,i,j,k')

M(x,y,z)=x-y*exp(x*z); N(x,y,z)=y^2*z; P(x,y,z)= z^2-y

curlA=((diff(P,y)-diff(N,z))*i+(diff(M,z)-diff(P,x))*j +(diff(N,x)-diff(M,y))*k)

curlA

Answer :  -j*x*y*e^(x*z) - (y^2 + 1)*i + k*e^(x*z)

Since curl, is not conservative.   ■

EXAMPLE 8

Prove that is (a) conservative field, (b) find the potential function of , (c) find the work done in moving an object in this field from to .

Solution.  (a) From the fact that curl ,

is conservative if ,

Hence is conservative.

(b) Let be a potential function such of . Then comparing the components of , and , we get

,

,

.

Integrating with respect to ,

.

Differentiating with respect to and using

.

Integrating with respect to

.

Substituting in

.

Differentiating with respect to and using

Integrating with respect to

.

Substituting in

.

(c) Work done

.

http://matrix.skku.ac.kr/cal-lab/Sec15-5-Exm-8.html

var('x,y,z,i,j,k')

M=y^2*cos(x)+z^3; N=2*y*sin(x)-4; P= 3*x*z^2+2

curlF=((diff(P,y)-diff(N,z))*i+(diff(M,z)-diff(P,x))*j +(diff(N,x)-diff(M,y))*k)

print curlF

print bool(curlF==0)

var('x,y,z,c')

fx=y^2*cos(x)+z^3; fy=2*y*sin(x)-4; fz= 3*x*z^2+2

f_1=integral(fx,x)    # +g(y,z)

f_1y=diff(f_1,y)      # fy= diff(f_1,y)+gy(y,z)

gy(y,z)=fy-diff(f_1,y)

g(y,z)=integral(gy(y,z),y) # +h(z)

f_2=f_1+g(y,z) #+h(z)

f_2z=diff(f_2,z)     #fz=diff(f_2,z)+ hz(z)

hz(z)=fz-diff(f_2,z)

h(z)=integral(hz(z),z)

f=f_2+h(z)+c

f   # This is a potential function for F

Answer :   x*z^3 + y^2*sin(x) + c - 4*y + 2*z

var('x,y,z,c')

f(x,y,z)=x*z^3 + y^2*sin(x) + c - 4*y + 2*z

x0=0;y0=1;z0=-1

x1=pi/2;y1=-1;z1=2

workdone=f(x1,y1,z1)-f(x0,y0,z0)

workdone

You will have the same answer : 4*pi+15.

15.5 EXERCISES (Curl and Divergence)

1. Find (a) the curl and (b) the divergence of the vector field.

Solution. (a) curl :

(b) div :

Solution. (a) curl :   ,  (b) div :

Solution.   (a) curl :    (b) div :

Solution. (a) curl :    (b) div :

Solution. (a) curl :    (b) div :

Solution. (a) curl :

(b) div :

Solution. (a) curl :     (b) div :

Solution. (a) curl :     (b) div :

2. Let be a scalar function and a vector field. State whether each expression is meaningful. If not, explain why. If so, state whether it is a scalar function or a vector field.

Solution. (a)Not meaningful, curl must take a vector field.

(b) Meaningful, vector field

(c) Meaningful, scalar function

(d) Meaningful, vector field

(e) Not meaningful

(f) Meaningful, vector field

(g) Meaningful, scalar function

(h) Not meaningful

(i) Not meaningful

(j) Not meaningful

(k) Not meaningful

(l) Meaningful, scalar function

3. Is there a vector field on such that curl ? Explain.

Solution. No, curl.

4. Is there a vector field on such that curl ? Explain.

Solution. No, curl.

5. (a) Let be a differentiable vector field with div. Define the vector field by

, , and

. Prove that .

(b)Let . Find such that .

6. Evaluate the line integral , where and is the curve given by , , .

Solution. Let . We then observe that , . Note first that direct computations show that , which implies that there exists a scalar function with . We then have the following equations: , , . Solving these equations, one can see that , where is a constant. Thus, due to the Fundamental Theorem for line integral,

.

7. Find .

Solution.

x,y,z,i,j,k=var('x,y,z,i,j,k')

P(x,y,z) = exp(x*z);

Q(x,y,z) = x*y^2*z;

R(x,y,z) = -y^2+z^2;

A(x,y,z) = P*i+Q*j+R*k;

A(x,y,z)

Answer :  j*x*y^2*z - (y^2 - z^2)*k + i*e^(x*z)

divA = diff(P,x)+diff(Q,y)+diff(R,z)

divA(x,y,z)

Answer : 2*x*y*z + z*e^(x*z) + 2*z

8. If  .

a.Prove that the line integral is independent of the curve joining two given points and .

b. Show that there exists a scalar function such that and find .

c. Also find the work done in moving an object from to .

9. Let and .

Then verify the following identities:

(a)

(b) curl

(c)

(d)

15.6 Surface and Area

The curves in dimensional space can be defined by a function , an equation , or parametrically by equations , . A curve described by a continuous function can be parameterized by letting so that parametric equations are , . We have two variables to parameterize a surface in defined by a function of two variables . Letting and , then parametric equations for are , and .

Establishing the Area of the Surface

Let be a smooth surface given by the vector valued function

,

where . (See Figure 1.) The functions , and are assumed to have continuous partial derivatives with respect to and . The rectangular region in the -plane is partitioned into rectangles, with sides of length and that are ordered in some convenient way, for . The rectangle corresponds to a curved path on the surface , with area . Let be the lower left corner point of . The other corners can be expressed as , , and so the area of is . Two of the edges of can be approximated by the vectors

.

As seen in Figure 1 these vectors actually form two of the edges of parallelogram lying in the tangent plane at . The area of the parallelogram approximates the area of

.

The Riemann sum

.

gives an approximation of the area of that part of surface corresponding to the point in .

The surface area of is

.

Then we can define the surface area of as the following definition.

Figure 1

DEFINITION1

Let be a continuous function on a smooth surface given a parameterization

where .

The tangent vectors

and

are continuous on and the normal vector is nonzero on . Then the surface area of is

.

Note that we will see in section 15.7.

EXAMPLE 1

Find the surface areas of the following surfaces:

(a) A cylinder with radius and height (excluding the circular ends).

(b) A sphere of radius .

Solution. The critical step is finding the normal vector . It needs to be done only once for any given surface.

(a) As shown before, a parametric description of the cylinder is

,

where and . A normal vector is

.

Notice that the normal vector points outward from the cylinder, away from the -axis. (See Figure 2.)

Figure 2

It now follows that

.

By ,  the surface area of the cylinder is

.

http://matrix.skku.ac.kr/cal-lab/Sec15-6-Exm-5.html

var('x,y,z,u,v,r,h')

assume(r>0)

x=r*cos(u);y=r*sin(u);z=v

r_u=vector([diff(x,u),diff(y,u),diff(z,u)])

r_v=vector([diff(x,v),diff(y,v),diff(z,v)])

n=r_u.cross_product(r_v)

print n

integrand=n.norm()

integral(integral(integrand,v,0,h),u,0,2*pi)

(b) A parametric description of the sphere is

, where and .

(See Figure 3.)

Figure 3

A normal vector is

.

Computing requires several steps. However, the needed result is quite simple: and the normal vector points outward from the surface of sphere. By , the surface area of the sphere is

.             ■

EXAMPLE 2

Find the area of the cone , where , .

Solution. The surface is an upper portion of the corn. First we compute

,

and then form the cross product

The magnitude of the vector is

.

Thus, the area is

.

Surface Area of Graph Surfaces

Any surface described by can be parameterized by the parametric equations

, , .

We can parameterize by setting

and so we obtain and .

Since the normal vector is

,

the length of is

.

Therefore, the area of surface is

.

EXAMPLE 3

Find the area of the surface , , .

Solution. Let , and with and . Since the partial derivatives of are ,

by , the area of the surface is

.  ■

EXAMPLE 4

Find the surface area of the portion of the surface , which is contained within the cylinder .

Solution.

var('x, y, z')

p1=plot3d(1/2*x*y, (x, -2, 2), (y, -2, 2), color='red', opacity=0.5)

p2=implicit_plot3d(x^2+y^2-4==0, (x, -2, 2), (y, -2, 2), (z,-2, 2), color='blue', opacity=0.5)

show(p1+p2)

Figure 4

Let .

Then

By the polar coordinate system, letting and with and ,

.

var('r, theta')

1/2*integral(integral(sqrt(4+r^2)*r, r, 0, 2), theta, 0, 2*pi)

Answer :  1/3*pi*(16*sqrt(2) - 8)    ■

Oriented Surfaces

A surface is orientable if it has two sides. Then one can orient the surface by choosing one side to be positive side. However some orientable surfaces are not orientable because they have only one side. One classical example is called the Möbius Strip. The Möbius Strip is not orientable because this surface only has one side (see Figure 5.). The Möbius Strip, named after the German mathematician August Möbius (1790-1868).

Figure 5 Möbius strip

More precisely, we say a smooth surface is an oriented surface if there exists a continuous unit normal function defined at each point on the surface. The unit normal vector is called an orientation of the surface. Note that if is an orientation of , then is also an orientation of in the opposite side. (See Figure 6 (a).)

A surface defined by has an upward orientation when the unit normals are directed upward, and has downward orientation when the unit normals are directed downward. (See Figure 6 (b), (c).)

Figure 6

In this case we define a new function or depending on the orientation of . If a smooth surface is defined by , then the unit normal vector on the surface is

or .

If a smooth surface is defined by a parametric equation

,

then the unit normal vector on the surface is

,

and the opposite orientation is given by

EXAMPLE 5

Find two orientations of the surface of the sphere

().

Solution. Method 1. Let .

We have the gradient of and the norm of the gradient of as follow;

and

.

Thus, by , two orientations of the surface are

and

.

The vector field defines on outward orientation, whereas defines an inward orientation from the sphere.

Method 2. This sphere can be parameterize by parametric equations

, ,

.

Since and , the unit normal vector of the sphere is

.

Also we have the opposite orientation

.   ■

15.6 EXERCISES (Surface Area)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-6-Sol.htm

1-3. Find a parametric representation of the surface.

1.

Solution. Let the parameters and ,

we find that parametric surface is given by equations , and .

Therefore, the plane is described by the vector function  .

2.

Solution. Let the parameters and , we find that parametric surface is given by equations , and .

Therefore, the plane is described by the vector function

.

3. The portion from to .

4-5. Find an equation of the tangent plane at the point on the surface corresponding to the given parameter values.

4. , , , , .

Solution. At and , the point on the surface is . The surface is defined by the vector function

, then

,

and .

The normal vector to the surface at and . An equation of the tangent plane at that point is or .

5. ; , .

6-8. Calculate the area of the surface.

6. The surface with parametric equations ,

, , , .

Solution. , , and then form the cross product

The area is

.

7. The part of the plane that lines in the first octant.

8. The part of the plane tha lines inside the cylinder .

15.7 Surface Integrals

As we did with line integrals we now need to move on to surface integrals of vector fields.

Surface Integral

Figure 1

We think about integrating functions over some surface in three-dimensional space. The region will lie above some region that lies in the -plane. Also note that we could just as easily look at a surface that was in front of some region in the -plane or the -plane.

If the surface is given by and is a continuous function, then the surface integral is

.

The surface integral is similar to areas of surfaces in section 15.6.

Next we have the definition of surface integral for parametric surface.

DEFINITION1

Let be a smooth surface given the vector valued function

,

and be a continuous function.

Then the surface integral of the function over is

.

If , the integral is the surface area of ;

.

EXAMPLE 1

Evaluate the surface integral where is the portion of the plane that lies in front of the -plane.

Solution. For using , we need to write the equation of the surface in the form . This is . (See Figure 2)

Figure 2

Next we need to determine just what is. Then is the projection of on -plane, i.e., .

Since and

.    ■

var('x,y,z')

implicit_plot3d(x+y+z==1,(x,0,1.2),(y,0,1.2),(z,0,1.2))

EXAMPLE 2

Evaluate , where is the sphere .

Solution. For the sphere, the parametric equations are

, and , for and .

The vector function is

.

Since the normal vector is

,

.

Then  .

http://matrix.skku.ac.kr/cal-lab/Sec15-7-Exm-2.html

var('x,y,z,u,v, n, nn')

assume( sin(u)>0 )

x=3*sin(u)*cos(v);y=3*sin(u)*sin(v);z=3*cos(u)

f=sqrt(x^2+y^2+z^2).simplify_trig()

r_u=vector([diff(x,u),diff(y,u),diff(z,u)])

r_v=vector([diff(x,v),diff(y,v),diff(z,v)])

n=r_u.cross_product(r_v)

nn=(n.norm().simplify_full()).simplify_trig()

t=f*nn

integral(integral(27*sin(u), u, 0, pi), v, 0, 2*pi)

The concept of surface integral is a generalization of a double integral taken over a plane region . In a surface integral is integrated over a curved surface.

Let be a two-sided surface with one side of taken arbitrarily as the positive side (the outer side if is closed) in Figure 3. A unit normal at any point of the positive side of is known as the positive outward  unit normal.

In the , the equation of a surface is given by with the unit normal .

When is represented in parametric form as with the two parameters and varying in a region of -plane, then the unit normal to at is given by where .

Figure 3

The surface integral of a continuous vector function taken over an oriented surface with unit normal is defined as

.

This integral is called the flux of across .

If is a parameterization of the surface , then

.

where is the parameter domain.

Let be the surface of the function on in -plane and    be a vector field.

Since , . Thus

.

Flux

Suppose the velocity of a fluid in is described by the vector field . Let be a surface in . The flux across is the volume of fluid crossing per unit time. Figure 4 below shows a surface and the vector field at various points on the surface.

Figure 4

What is the formula for the flux? Let us consider the flux through an infinitesimal piece of the surface with area above the point in the -plane. (The surface is denoted by the dotted region.) Let denote the unit normal vector to the surface. Let us suppose that the velocity vector of the fluid is . The normal component is a scalar. Then the flux of represents the total quantity of fluid flowing in unit time through (across) the surface in the positive direction. The flux of across is given by the surface integral

Flux of across .

EXAMPLE 3

Find the flux of the vector field across the sphere .

Solution. Using the spherical coordinate system, the parametric equations

with and .

Then and

.

.    ■

Evaluation of a Surface Integral

A surface integral is evaluated by reducing it to a double integral. Let be the projection of onto the -plane.

Then, and

.

where is the unit outward drawn normal to . The double integral on the right in , over the plane region is evaluated as an iterated integral.

In a similar way the surface integral can be evaluated by projecting onto the -plane as and -plane as as follows

,

.

EXAMPLE 4

Evaluate the surface integral where and is the surface bounded by the cylinder , , , and . (See Figure 5.)

Figure 5

Solution. The entire surface consists of 5 surfaces: is the curved (lateral) surface of the cylinder , is the parallelogram , is the parallelogram , is the quarter of circle , is the quarter of circle . Then

.

We will use - to evaluate the surface integrals of , .

On the curved (lateral) surface of the cylinder, since the unit normal vector is

, and .

And the projection of onto -plane is . Then

.

On , since and , and . Then

.

On , since and , and . Then

.

On , since and , and . Then

.

On , since and , and . Then

.

Thus the required surface integral is

15.7 EXERCISES (Surface Integrals)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-7-Sol.htm

1.Evaluate the surface integral , where is the surface defined by the vector function

, where , .

Solution. The graph of is called sphere.

Substituting , and into the integrand and we have

.

and

.

2. Evaluate , where is the surface with parametric equations , and , .

3. Evaluate, where is the part of the plane that lies in the first octant.

4. Evaluate, where is the hemisphere .

5. Evaluate, where is the portion of the paraboloid between and .

6. Evaluate where and is that part of the plane which is  located in the first octant.

Solution.   Here can be thought of as a graph surface of a function . The unit normal to this surface is given by

.

Thus .

Hence .

7. If is the entire surface of the cube bounded by , , , , , and and then evaluate .

Solution. Here can be spilt into six surfaces , , , as shown in the figure. It is easy to see that normal to , , , are , , , , , respectively.

Therefore .

8. Let be the surface of the cylinder included in the first octant between and .

Evaluate where .

Solution. We spilt into five pieces , , , as shown in the figure.

,

Therefore .

9. Find the surface integral over the parallelepiped , , , , , when .

Solution.

,

,

,

,

,

.

Therefore .

10. If is the surface of the sphere and , evaluate .

Solution. The sphere can be given a parameterization

11. Evaluate .

Here and is the boundary of the solid surrounded by and .

Solution.

1) : There is a graph surface of on .

We have , , and .

Hence by ,

2) :

3) :  .

15.8 Stokes’ Theorem

Sir George Gabriel Stokes (1819-1903) Irish mathematician

We introduce another important theorem of vector calculus, namely Stokes’ Theorem. Stokes’ Theorem can be thought of as an extension of Green’s Theorem. Green’s Theorem relates the integral over a plane region to a line integral on the boundary of (with positive orientation). Stokes’ Theorem on the other hands relates a surface integral over an orientable surface to a line integral over the boundary of . The orientation on is taken as positive.

THEOREM 1  Stokes Theorem

Let be a vector field whose components have continuous second order partial derivatives in a domain containing a surface bounded by a simple closed curve . Then

where is traversed in the positive direction.

We shall prove this theorem for special class of surfaces namely a graph surface. A general proof of this theorem is beyond our scope.

Proof.  See Figure 1.

Figure 1

Let be a surface as , . Then is a position vector of .

Since   curl

and with ,

.

Let be a closed curve of and         be a vector field in -plane and be a position vector of . Since ,

.

Let be a closed curve of . Then

.

Thus

.

Let and .

By Green Theorem, can be changed the following:

.

By and , .                                ■

Stokes’ Theorem can be expressed by

where is the surface of that is a position vector and is the projection of on -plane. And Stokes’ Theorem in rectangular form is

where , and , , are angles made by the normal with the unit vectors , and .

Green’s Theorem in the plane is a special case of Stoke’s Theorem.

Suppose is a surface lying in the -plane, with upward unit normal vector . Let be the positive oriented boundary of .

Then which is nothing but the vector version of Green’s Theorem.

The circulation of around a closed curve is given by the line integral where represents the velocity of a fluid circulation. This has applications in fluid mechanics and aerodynamics.

EXAMPLE 1

Let be that the part of the plane in the first octant, oriented with the upward-pointing normal and let be its boundary, oriented counterclockwise. If , verify Stokes’ Theorem by computing both and . (See Figure 2.)

Figure 2

Solution. Stokes’ Theorem states that

.

First we will find .

Here is the curve consisting of , and . So

.

On , a simple parameterization of this line segment is

where .

Thus . In terms of this parameterization,

the vector field becomes

.

Then .

Thus .

Similarly, on , it's parameterized by , so

.

Thus .

Similarly, on , it's parameterized by , so

.

Thus .

Hence, .

Next we will find . This triangular surface is a graph (using ) over the triangle in the first quadrant of the -plane. Thus we can parameterize it as

for .

Since , we get

curl .

Similarly,

,

Then

.    ■

EXAMPLE 2

Evaluate where and is the portion of the plane that lies inside the sphere .

Solution.

var('x, y, z')

p1=implicit_plot3d(x^2+y^2-1==0, (x, -2, 2), (y, -2, 2), (z, 0, 6), color='blue', opacity=0.3)

p2=implicit_plot3d(x+y+z==4, (x, -2, 2), (y, -2, 2), (z,0, 6), color='red', opacity=0.5)

show(p1+p2)

Figure 3

In Stokes’ Theorem, we will use

curl ,

Since ,

.

Since is a unit circle in -plane,

.   ■

Stokes theorem relates the surface integral of the curl of a vector field over a surface in Euclidean three-space to the line integral of the vector field over its boundary :

This Divergence theorem, fundamental theorem of calculus, and Green’s Theorem are simply speical cases of the general formulation of the Stokes theorem.

15.8 EXERCISES (Stokes’ Theorem)

1.Evaluate using Stokes’ Theorem, given that is the circle: that lies inside the cylinder and above the -plane.

Solution. Note that the curve of intersection is the circle at the plane .

http://matrix.skku.ac.kr/cal-lab/cal-15-8-1.html

var('x, y, z, t')

p1 = implicit_plot3d(x^2+y^2+z^2==4, (x, -2, 2), (y, -2, 2), (z, 0, 2), opacity=0.2, color="red", mesh=True);

p2 = implicit_plot3d(x^2+y^2==1, (x, -2, 2), (y, -2, 2), (z, 0, sqrt(3)), opacity=0.5, color="blue", mesh=True);

p3 = plot3d(0, (x, -2, 2), (y, -2, 2), opacity=0.3, color="orange", mesh=True);

show(p1+p2+p3, aspect_ratio=1)

var('i, j, k')

r_t = vector(SR, [cos(t), sin(t), sqrt(3)]);

dr_t = diff(r_t, t);

F = vector(SR, [x, y, x*y]);

F_r_t = vector(SR, [cos(t), sin(t), cos(t)*sin(t)]);

print (F_r_t)

integral(F_r_t*dr_t, (t, 0, 2*pi))

2. Evaluate (a) directly (b) using Stokes’ Theorem where is the ellipse that intersect between and .

Solution. Let .

So and . Hence,

.

3. Verify Stokes’ Theorem for the vector field over an orientable surface which is the upper hemisphere and .

Solution. Note that the curve is , with positive orientation. Since curl , we have curl .

Next we compute where , .

Hence

.

Hence verified.

4. Verify Stoke’s Theorem for a vector field over an orientable surface which is the square in the - plane.

[Hint] is a square of .

Solution.

var('x,y,z,a')

M=x^2; N=x*y; P=0

curlF=vector([diff(P,y)-diff(N,z),diff(M,z)-diff(P,x),diff(N,x)-diff(M,y)])

print curlF

x=x;y=y;z=0

r_x=vector([diff(x,x),diff(y,x),diff(z,x)])

r_y=vector([diff(x,y),diff(y,y),diff(z,y)])

n=r_x.cross_product(r_y)

w=curlF.dot_product(n)

5. Verify Stokes’ Theorem for a vector field over a rectangle bounded by .

Solution.

var('x,y,z,a')

assume(a>0)

M=x^2+y^2; N=-2*x*y; P=0

curlF=vector([diff(P,y)-diff(N,z),diff(M,z)-diff(P,x),diff(N,x)-diff(M,y)])

print curlF

x=x;y=y;z=0

r_x=vector([diff(x,x),diff(y,x),diff(z,x)])

r_y=vector([diff(x,y),diff(y,y),diff(z,y)])

n=r_x.cross_product(r_y)

w=curlF.dot_product(n)

6. Evaluate , where and is the curve which is the intersection of and . ( is upward anticlockwise).

Solution. curl and the curve is a boundary of on . Using Strokes' Theorem and the ploar coordinate system,

curl .

7. We consider the vector field and the curve which is the boundary of the triangle with vertices . Compute the work done by the force field in moving a particle along the curve . (First, the particle goes from to , and goes from to , finally goes from back to ).

Solution. Consider the region . The is the boundary of . Since , and ,

Using Stokes’ Theorem, we have

.

8. Evaluate . Here and is a triangle with vertices  .

Solution. curl .

The surface bounded by is and this can be represent a parametric vector  . So we get . And the domain is the projection of the surface on -plane as .

curl .

15.9 Divergence Theorem

Carl Friedrich Gauss (1777–1855), German mathematician. Painted by Christian Albrecht Jensen

This section deals with the third important theorem, called the Gauss Divergence Theorem. It is an extension of the vector version of Green’s Theorem to vector fields in and relates the integral of a derivative of function (div of a vector field ) over a region to the integral of the function over the boundary of the region. The region in this case is considered as a solid region and its boundary is a surface. We make certain assumptions about the region and its boundary.

We assume that is a closed surface, that is encloses a solid region in . For example, spheres, cubes and ellipsoids are closed surfaces whereas planes, paraboloids, etc are not closed regions. We further assume that has positive orientation. That is, the unit normal vector is directed  outward from .

THEOREM 1  Gauss Divergence Theorem

Let be a closed surface in which is the boundary of a solid region . Let be a vector field defined on some open set containing . Suppose the components of have continuous first order partial derivatives and is an unit normal vector that is directed outward from . Then

div .

Proof.  We prove this theorem for a particular type of . Assume is a surface such that any line parallel to coordinate axes meets in at most two points. Let and be the lower (below) and upper (top) portions of having equations and and having and as normals respectively (See Figure 1).

Figure 1

Let be the projection of the surface on the -plane. If , then the result of Gauss Divergence Theorem in component form is

.

Thus it is sufficient to prove

,   ,

.

Consider the next equation.

.

Since for the upper surface , while for the lower surface , , then

.

Similarly, projecting on to the -plane and -plane we have

,

.

Adding , and , we get the required result .

Gauss Divergence Theorem (GDT) transforms volume integrals to surface integrals and vice versa.

GDT in rectangular form can be written

where , , . Here , , are the angles which makes with the positive , , axes.

GDT is also known as “Green’s Theorem in space” because the GDT generalizes the “Green’s Theorem in the plane” by replacing the (plane) region and its closed boundary (curve) by a (space) region and its closed boundary (the surface).

The left hand side of is referred as the flux of through the surface . Thus Gauss Divergence Theorem says that under the certain conditions, the flux of across the surface is equal to the triple integral of the divergence of .

When velocity of a fluid then the GDT has the following physical interpretation:

 Volume of fluid emerging (diverging) from a closed    surface in unit time Volume of fluid supplied from within volume in unit time

Let us verify the Gauss Divergence Theorem in the next examples.

EXAMPLE 1

Verify of the Gauss Divergence Theorem for taken over the region in the first octant bounded by the cylinder and the plane . (See Figure 2.)

Figure 2

Solution. Since and

,

then

.

The entire surface consists of five surfaces , , , , . So

.

We will use - in section 15.7 to find the followings.

On : , since and , . Then

.

On : , since and , and .

Then

.

On : , since and , . Then

On : , since and , . Then

.

On : of the cylinder: , since the unit normal is

,

and

.

Projecting the surface on the -plane

.

Thus . Therefore, the Gauss Divergence Theorem is verified.

EXAMPLE 2

Find the flux of a vector field over the sphere using the GDT.

Solution. Note that div . Thus by GDT, we have flux of across

div

Volume of the Sphere .

EXAMPLE 3

Find the surface integral , where over the surface which is bounded by and .

Figure 3

Solution. Note that div . Thus by the GDT, we have

By the polar coordinate system, letting , , then and . Thus

.  ■

http://matrix.skku.ac.kr/cal-lab/cal-15-9-Exam-3.html

var('x,y,z');

def Div(F):

assert(len(F) == 3)

return (diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

F=vector([x^3,y^3,z^2])

Div(F)

Answer :  3*x^2 + 3*y^2 + 2*z

var('t,r')

x=r*cos(t);y=r*sin(t) # polar coordinate system

f=3*x^2 + 3*y^2 + 2*z

integral(integral(integral(f*r,r,0,3),t,0,2*pi),z,0,2)

In the next example we have to find the volume of a region bounded by a surface by the Gauss Divergence Theorem.

EXAMPLE 4

Find the volumd of a region bounded by a surface .

Solution. By Gauss’ Divergence Theorem

.

Choose , so that , with this reduces to

Volume.

Similarly by taking and , we get

and   or .   ■

Divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region.

The divergence theorem is an important result for the mathematics of engineering, in particular in electrostatics and fluid dynamics.

15.9 EXERCISES (Divergence Theorem)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-15-9-Sol.htm

1-4. Using the Divergence Theorem, evaluate the surface integral

1. and the boundary of a region is the sphere .

Solution. Define “Div” function

var('x,y,z')

def Div(F):

assert(len(F)==3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

F=vector([y*z,z*x,x*y])

Div(F)

Since divergence is 0, .

2. and is a closed surface consisting of the circular cylinder and the circular disks and .

Solution.

var('x,y,z')

def Div(F):

assert(len(F)==3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

F=vector([x^3,x^2*y,x*y])

Div(F)

var('a,b,t,r')

assume(a>0,b>0)

x=r*cos(t);y=r*sin(t) # polar coordinate system

f=4*x^2

integral(integral(integral(f*r,r,0,a),t,0,2*pi),z,0,b)

Then .

3. and is the surface bounded by , , .

Solution.

var('x,y,z')

def Div(F):

assert(len(F)==3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

F=vector([sin(x),2-y*cos(x),0])

Div(F)

Since divergence is 0, .

4. and is the cube of side .

http://matrix.skku.ac.kr/cal-lab/cal-15-6-4.html

Solution.

var('x,y,z,b')

assume(b>0)

def Div(F):

assert(len(F)==3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

F=vector([x^2-y*z,-2*x^2*y,z])

integrand=Div(F)

integral(integral(integral(integrand,x,0,b),y,0,b),z,0,b)

Answer:  -1/3*(2*b^3 - 3*b^2 - 3*b)*b^2

5-13. Verify the Gauss Divergence Theorem for :

5. taken over the region bounded by , and .

Solution.

var('x,y,z,t')

p1 = implicit_plot3d(x^2+y^2==4, (x,-2,2), (y, -2,2), (z, -5,5), opacity=0.2, color="red", mesh=True);

p2 = implicit_plot3d(z==0, (x,-2,2), (y, -2,2),(z, -5,5), opacity=0.3, color="blue", mesh=True);

p3 = implicit_plot3d(z==3, (x,-2,2), (y, -2,2), (z, -5,5),   opacity=0.5, color="orange", mesh=True);

show(p1+p2+p3, aspect_ratio=1)

var('x,y,z,t,r')

def Div(F):

assert(len(F)==3)

return (diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

f=Div([4*x,-2*y^2,z^2])

g(r,t,z)=f.subs(x=2*cos(t),y=2*sin(t),z=z)

integral(integral(integral(g*r,r,0,2),t,0,2*pi),z,0,3)

, , ,

, .

Therefore .

6. taken over the entire surface of the cube , , .

Solution.

,

,

,

(Since on ),

,

Therefore .

Since div, div.

7. taken over the entire surface of the sphere of radius and centered at the origin.

Solution.

var('a,b,c,x,y,z')

def Div(F):

assert(len(F) == 3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[1],z))

Div([a*x,b*y,c*z])

So, integral of a+b+c over the entire surface of the sphere of radius is . Compute the second part, and show they are equal.

8. and is the total surface of the rectangular parallelepiped bounded by the coordinate planes and , , .

Solution.

var('x,y,z,t')

p1 = implicit_plot3d(x==1, (x,0,5), (y, 0,5), (z, 0,5), opacity=0.2, color="red", mesh=True);

p2 = implicit_plot3d(y==2, (x,0,5), (y, 0,5), (z, 0,5), opacity=0.3, color="blue", mesh=True);

p3 = implicit_plot3d(z==3, (x,0,5), (y, 0,5),(z, 0,5),   opacity=0.5, color="orange", mesh=True);

show(p1+p2+p3, aspect_ratio=1)

def Div(F):

assert(len(F) == 3)

return

(diff(F[0],x)+diff(F[1],y)+diff(F[1],z))

d10=Div([2*x*y,y*z^2,x*z])

integral(integral(integral(d10,x,0,1),y,0,2),z,0,3)

Compute the second part, and show they are equal.

9. over the upper half of the sphere .

Solution.   A parameterization of sphere is

,

,

, .

Hence (See the figure).

,

.

Therefore .

Compute the second part, and show they are equal.

10. taken over the rectangular parallelepiped bounded by the coordinate planes and , and .

Solution. .

11. taken over the surface of the ellipsoid .

Solution. Let be a parameterization of .

,

.

Hence

.

Compute the second part, and show they are equal.

12. taken over the upper half of the unit sphere .

Solution.

,

,

.

Therefore .

Compute the second part, and show they are equal.

13. taken over the closed region of the cylinder , bounded by the planes and

Solution. ,

.

,

Therefore .

Compute the second part, and show they are equal.

14. (a)  Prove Green’s first identity :

(b) Let and

. Find .

15. Let be a solid surrounded by and and vector field . Find the flux of , that is,

Solution. div,

div

16. Evaluate .

Here , and is the boundary of the solid surrounded by , .

Solution. div.

Let .

.

17. Let be a surface between and and . Find a flux .

Solution. div.

div

18. Evaluate the surface integral , where

and is the part of surface of the paraboloid above -plane.

Solution. Let and be the region bounded by and .

Then, with the aid of the Divergence Theorem,

.

Since and on , we can see that . On the other hand, , and therefore, we have = volume of , which can be computed as follows:

.

19. Let .

Evaluate , where is the part of the sphere , .

Solution. Note that ,

Using the Divergence Theorem and polar coordinates , , where

,

we compute .

div

.

.

:

.

20. If , evaluate over the volume of a cube of side .

21. Evaluate over the solid region of the sphere when where , , are constants.

Calculus