Calculus

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Chapter 14. Multiple Integrals

14.1 Double Integrals http://youtu.be/jZ2pAmPZYOE

문제풀이 by 이인행 http://youtu.be/w8g9fgcEP4A

14.2 Double Integrals in Polar Coordinates http://youtu.be/olQgihl5aZg

문제풀이 by 이지석 http://youtu.be/jpsObxtZ50A

14.3 Surface Area http://youtu.be/p9R0TTLfBzk

14.4 Cylindrical Coordinates and Spherical Coordinates

문제풀이 by 최양현 http://youtu.be/F9u6pMubVRs

14.5 Triple Integrals http://youtu.be/r1tzH9Ibbqk

문제풀이 by 이인행 http://youtu.be/C-uPM3km96k

14.6 Triple Integrals in Cylindrical and Spherical Coordinates

14.7 Change of Variables in Multiple Integrals http://youtu.be/INn-bkgXYNg

14.1 Double Integrals

In this section, we consider the integration of the function on the domain in the -plane, which is called a “double integral.” This will be used to define the volume of the region under the graph of a function . We do not require , but if assumes negative values, we shall not interpret the double integral as a "volume". At first, we define the double integral over a rectangle by introducing the limit of Riemann sums and then we consider the double integral over the general domain in the -plane. Next, we deal with Fubini's Theorem which allows us to compute the double integral by computing iterated integrals over the interval. Finally, we introduce some applications of the double integral, for example, center of mass and total mass when the density function is given.

To begin, let us establish the notation for partitions and sums.

Figure 1

Consider a rectangle in . A partition of is two collections of equally spaced points and such that

,

and

(), ().

Let be the rectangle and let be any sample point in . If , then we form the following Riemann sum for :

where is the area of the subrectangle .

Then we can consider the limit of , as and tend to .

If is a bounded and continuous function on the rectangle , then one can prove that always exists and the limit is independent of the choice of the sample point . This leads to the following definition.

Figure 2

DEFINITION1

Let be a bounded real-valued function on a rectangle . If the Riemann sum defined above converges to a limit , independent of the choice of the sample point , then we say that is integrable over and we write

or

for the limit .

We have considered double integral over by dividing into equal-sized rectangles. However, this is not necessary. One can arrive at the same definition by arbitrary partitions.

If on a rectangle and is integrable then represents the volume of the solid that lies above and below the surface .

EXAMPLE1

Let be defined by . Show that is integrable and hence find .

Solution. Choose a subrectangle , then we form the Riemann sum

where .

Now

.

From the formula , we have

.

By the Squeezing Theorem, we have

Thus is integrable on and . ■

EXAMPLE2

Let on . Show that is integrable on and hence find its integral.

Solution. Choose a subrectangle , then we form the Riemann sum

,

where and .

Now

.

From the formula and , we have

.

By the Squeezing Theorem, we have

Thus is integrable on and . ■

Properties of a Double Integral

From the definition of the double integral, one can guess that double integrals share properties with the single variable definite integral. These properties are similar to proof in case of single variables.

THEOREM 2

Let and be integrable functions on the rectangle , and let be a constant. Then and are integrable on and

(i) (Linearity)

.

(ii) (Homogeneity) .

(iii) (Monotonicity) If , for all then,

.

(iv) If for all in , then

.

(v) Furthermore, if is continuous on , then there exists at least one point in such that

.

The property (v) of Theorem 2 is called the “mean value theorem for double integrals”.

All the above properties can be proved by using the definition of the integrals and the limit of sums, the limit theorems and the definition of continuity.

Double Integral over a General Domain

We want to extend the definition of a double integral to functions which are defined on more general regions. We restrict our attention to the “elementary domains”. We say the following two types of domains are “elementary domains”.

Ⅰ.Suppose that we are given two continuous functions that satisfy for all .

Let . (See Figure 3.)

In this case is called a typeⅠ domain.

Ⅱ.Suppose that we are given two continuous functions that satisfy for all .

Let . (See Figure 4.)

We call such a domain a type II domain.

The curves or straight lines that bound the region constitute the boundary of , denoted by .

Note that there are some regions which can be expressed as both type in Ⅰ and Ⅱ.

For example, the rectangle and disk are of both types.

Figure 3 Figure 4

Once we define the double integral over elementary domains, then this definition can be extended to an arbitrary domains in the following ways.

Let be a continuous and bounded function on an arbitrary domain in . Let be a rectangle which contains . Now we extend on to a function defined on as follows:

Then is bounded and continuous except on the boundary of . Now we can define Riemann sum for defined on . If and are very large, then only negligible number of subrectangles meet (because is elementary domain). Thus if is Riemann integrable on , then we say that is Riemann integrable on . (This is not a rigorous proof, but a rigorous proof depends on this idea.)

Thus for a continuous on , we can define the double integral on as follows:

.

We defined the double integral over the domain by choosing a rectangle containing . But intuitively, the value of the double integral is independent of the choice of the rectangle because outside , is zero.

Iterated Integral

As in the case of one variable, evaluation a double integral from the definition is not an easy task. However, double integrals under certain conditions can be evaluated by two single variable integrals.

Let is an integrable function on a rectangle . If we fix in , then is a single variable function of on . Therefore, we can define as a Riemann integral of one variable if it exists.

Note that is a function of . We can denote by . Now we can integrate on if it exists and we get

.

The right hand side of is called an iterated integral. It is standard practice to omit square brackets in .

Thus

means that first we integrate with respect to in and then with respect to in .

Similarly one can define

.

It turns out that in most of the cases the two iterated integrals and are equal.

The following theorem (due to Guido Fubini's Theorem, 1907) gives a way to evaluate double integrals. http://en.wikipedia.org/wiki/Fubini's_theorem

THEOREM 3 Fubini’s Theorem

Let be a continuous function defined on the a rectangle . Then

Theorem 3 can be proved using the mean value theorem for integrals in the small subrectangles. We omit the details of the proof.

Although Theorem 3 is named after Guido Fubini, it was certainly known to Cauchy and his contemporaries for continuous functions. Fubini proved that the above equality is true for general functions. The original form of the Fubini’s Theorem is beyond the scope of this book.

EXAMPLE3

Evaluate the following integrals.

(a) (b)

Solution. (a) We first integrate with respect to and then with respect to

.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-2.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x')

f(x) = 2 + 16*x

integral(f, x, 0, 1)

var('x, y')

f(x, y) = 1 + 4*x*y

integral(integral(f, y, 1, 3), x, 0, 1)

Answer : 10

(b) We first integrate with respect to and then with respect to

.

var('x, y')

f(x, y) = 1 + 4*x*y

integral(integral(f, x, 0, 1), y, 1, 3)

Answer : 10 ■

Thus we see that the two integrals in Example 3 are equal.

EXAMPLE 4

Evaluate .

Solution. By iterated integral Theorem, we have

.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-3.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x, y')

f(x, y) = x^3 + y^2

integral(integral(f, x, 0, 1), y, 0, 1)

Answer : 7/12 ■

If we consider the double integral over elementary domains, then we may rewrite the iterated integral.

Suppose is typeⅠ domain. (See Figure 3.)

.

and the extension is zero outside of . Then

.

If is a type Ⅱ domain. (See Figure 4.)

.

Let be an extension of on . Then

.

EXAMPLE 5

Evaluate the integral

where . (See Figure 5.)

Figure 5

Solution. By Fubini's Theorem, we have

var('x, y')

f(x, y) = x^3 + y^2

integral(integral(f, x, 0, y), y, 0, 1)

Answer : 3/10 ■

EXAMPLE 6

Evaluate , where is the region bounded by the parabola and the line .

Solution. The two curves intersect at , . (See Figure 6.)

Figure 6

We note that the region . So, we can write

. ■

EXAMPLE 7

Find the volume of the tetrahedron bounded by the coordinate planes and and .

Solution. We plot the tetrahedron in Figure 7.

Figure 7

We want to find the region in the -plane over which the tetrahedron lies. This region is obtained by projecting the plane onto the -plane.

The region is given by

. ■

If on , we can interpret the integral as the volume of the three-dimensional region between the graph of and the domain .

EXAMPLE 8

Evaluate .

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-5.html

Solution.

var('x, y')

f(x, y)=x*sin(x)*cos(y)

integral(integral(f, x, 0, pi), y, 0, pi/2)

Answer :

After interchanging the limit

integral(integral(f, y, 0, pi/2), x, 0, pi)

Answer : ■

Changing the Order of the Integration

Suppose that the domain can be written as a typeⅠ domain

and it can be also written as a type II domain

for some functions , and , .

Thus,

It is often useful to change the order of the iterated integrals. Many times the integral becomes easier to evaluate by interchanging the order of the iterated intervals.

EXAMPLE 9

Evaluate the integral .

Solution. It is not easy to evaluate the above integral in the above order.

Domain of the integral is

. (See Figure 8.)

This domain can also be written as

and (See Figure 9.)

Figure 8 Domain for integral Figure 9 Domain for integral

If we change the order of the integration, then we have

.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-6.html

var('x, y')

f(x, y)=1/(1+ y^2)^2

integral(integral(f, x, 0, y), y, 0, 1)

Answer : 1/4 ■

Suppose that and

.

Then Fubini’s Theorem gives

.

.

EXAMPLE 10

Evaluate the integral , where .

Solution. By Fubini’s Theorem,

.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-7.html

var('x, y')

f(x)=x

g(y)=y

integral(f, x, 0, 1)*integral(g, y, 0, 1)

Answer : 1/4 ■

In Figure 10 illustrates the division of a region into subregions and for which . The regions and can have no points in common except possibly on their common border which . Then

.

Figure 10 is the union of two disjoint regions and .

EXAMPLE 11

Evaluate , where is enclosed by and . (See Figure 11.)

Figure 11

Solution.

.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-Exm-8.html

var('x, y')

f(x, y)=x-1

g(x, y)=x-1

integral(integral(f, y, x, x^3), x, -1, 0) + integral(integral(g, y, x^3, x), x, 0, 1)

Answer : -1/2 ■

Areas of Regions by Doubles Integrals

Recall that the integral of the constant function over a region , it gives the area of :

Area.

EXAMPLE 12

Find the area of the region enclosed between the parabola and the line . (See Figure 12.)

Figure 12

Solution.

Area

. ■

Average Value of a Function in Plane

Recall that the average value of a function of defined on an interval is

.

Similarly, we define the average value of an integrable function of two variables defined on a region is

.

EXAMPLE 13

Find the average value of over the region .

Solution. The area of the region is .

The average value of is

. ■

EXAMPLE 14

Find the volume of the ellipsoidal solid .

Solution. When we slice the ellipsoidal solid by the plane that is through and parallel to plane, the intersection is an ellipse by

The area of this cross-section is

.

Recall that the area of ellipse is .

Hence the volume of the ellipsoidal solid is

■

Applications of the Double Integral

We have already seen an application of double integrals to special types of solids: computing volumes. This section will allow us to think about physical applications such as computing mass, electric charge, center of mass, and moment of inertia. We will see that these physical ideas are also important when applied to the joint probability density functions of two random variables.

If the region is a rectangle, the probability that lies between and and lies between and is

.

Here, is the joint probability density function that satisfy ,

.

Let be the density function at the point . Then the total mass of the lamina in can be written as

.

The center of mass can be defined by

and

.

Moments of inertia with respect to -axis and -axis, and , respectively, can be defined by the following formulas,

and .

The moment of inertia with respect to the origin is the sum of and as follows:

.

is called the polar moment of inertia.

EXAMPLE 15

Find the value of constant and for the following joint probability density function:

.

Solution. The constant is found from :

.

Therefore . (See Figure 13(a).)

Since ,

. (See Figure 13(b).) ■

(a) (b)

Figure 13

EXAMPLE 16

Find the mass and center of mass of a triangular lamina with vertices , and if the density function is .

Solution. See Figure 14. The mass of a triangular lamina is

.

Figure 14

Then and give

,

.

Therefore the center of mass is at the point . ■

EXAMPLE 17

Find the moments of inertia and if the density function is given over the domain

.

Solution. By the definition of the moment of inertia,

,

. ■

14.1 EXERCISES (Double Integrals)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-1-Sol.htm

1-3. Evaluate the given iterated integrals:

1.

Solution.

2.

Solution.

3.

Solution.

4-6. Sketch the region of integration in the iterative integrals.

4.

5.

6.

7-8. Evaluate the double integral over the given region that is bounded by the graphs of the given equations. Choose the appropriate order of integration.

7. , , .

8. , , , .

9. , is bounded the triangle with vertices , , .

Solution. Answer : 13/4

10. Change the order of integration.

(a)

(b)

Solution. (a)

(b)

.

11-12. Evaluate the following integrals by changing the order of the integration.

11. .

Solution.

12.

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-2.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x, y')

f(y)=2/(y^2 -2*y + 2005)

g(x)=f(x)-f(0)

integral(g, (x, 0, 1))

Answer : -log(2004) + log(2005)

13. Evaluate .

Solution.

.

14. Evaluate .

Solution.

15. Evaluate .

Solution.

16. Evaluate .

Solution.

17. Evaluate .

Solution.

var('x, y')

f(x, y)= (x-1)*(1+ e^(2*y))^1/2

integral(integral(f, y, 0, ln(x)), x, 1, 2) #

18. Find the volume of the solid

with .

19. Prove that

.

20-21. If the density function over the following domain , then find the center of mass, the moments of inertia and .

20.

21.

Let be the volume of the solid obtained by rotating

about -axis.

When we slice the solid by the plane which is through the point and perpendicular to -axis, its area of cross-section is . Thus the volume is

.

22. Find the volume of the solid obtained by rotating the region enclosed by and about -axis.

Solution.

23. Find and for the area enclosed by the loop of .

Solution.

where we have used the transformation . Then

24-25. Evaluate for the given function and region .

24. ,

.

25. ,

.

14.2 Double Integrals in Polar Coordinates

Double Integrals in Polar Coordinates

In section 10.3, we have seen that any point with cartesian coordinates can be converted into the polar coordinates by the following transformations

. and

Therefore, we would like to look at what happens to a double integral

, when and are transformed into polar coordinates.

One can easily show that the area of a small rectangle in polar coordinates gets transformed to in polar coordinates.

Area

,

where is the average radius . (See Figure 1-3.)

Figure 1 Figure 2 Figure 3

In particular, we have the following result which gives a method to transform a double integral in rectangular coordinates to polar coordinates.

THEOREM 1

Let be continuous functions on a polar rectangular region given by and where . Then

In particular, , the area of the region boundary , and , is

Area .

If is a region bounded by polar functions and with (See Figure 4.), then

.

Figure 4

On the other hand if is a region bounded by polar functions and with , then

.

EXAMPLE1

Find where

. (See Figure 5.)

Figure 5

Solution.

. ■

EXAMPLE2

Find by transforming this to polar coordinates.

Solution. Note that the region of integration is

.

Under the polar transform, it is mapped to

. (See Figure 6.)

Figure 6

. ■

EXAMPLE3

Find , where

.

Solution.

. ■

EXAMPLE4

Find the volume of the region above the surface and below the plane . (See Figure 7.)

Figure 7

Solution.

Note that the region of integration is

.

Under the polar transform, it is mapped to

.

. ■

EXAMPLE5

Find the volume of the solid inside the sphere and inside the cylinder . (See Figure 8.)

Figure 8

Solution. Let .

. ■

EXAMPLE6

Evaluate , where is the region in the first quadrant that is outside and inside the cardioid .

Solution. The two equations intersect at

. ■

EXAMPLE7

Show that .(Gaussian Integral)

http://mathworld.wolfram.com/GaussianIntegral.html

Solution. Let and be the regions in the first quadrant inside the disk , respectively. Let be rectangle. Then . (See Figure 9.)

Figure 9

Let , and .

Using Polar Coordinate we have

, and .

AS , and come to close at .

By Squeeze Theorem, comes to close at .

By Fubini’s Theorem,

.

Hence as , . ■

14.2 EXERCISES (Double Integrals in Polar Coordinates)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-2-Sol.htm

1. Find .

Solution. .

2. , where

Solution.

Since ,

.

3. Find the area inside the cardioid .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-2-2(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('v')

polar_plot(1+sin(v),(v, 0, 2*pi))

.

4. Find the area inside of .

Solution.

.

5. Find the volume of .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-2-4(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('u, v, x, y')

parametric_plot3d( (v*sin(u), v*cos(u), 4-v^2), (u, 0, 2*pi), (v ,0 ,2),color='pink' )

Note that , where is the disk: .

In polar coordinates can be written as . Hence

.

6. Find

Solution. .

7. Evaluate , where is the region bounded by , ,

and .

Solution. Using the polar coordinate system, the region is represented as follows:

and .

Here we used . Then the given integral becomes

.

8. Let .

Evaluate .

9. Let. Evaluate .

Solution.

Let and .

.

10. Find.

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-1-3.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x, y')

f(y)=integral(sqrt(1-y^2), y)

g(x)=f(sqrt(1-x^2))-f(0)

integral(g, (x, 0, 1))

Answer : 2/3

Let and .

.

14.3 Surface Area

Surface Area

Surface area is the total area of the faces and curved surfaces of a solid figure. A mathematical description of the surface area is considerably moras a sphere, are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration. In this section we apply the notion of double integrals to find the area of a surface. We consider a surface with equation , namely the graph surface. Let the part of surface area be , and it's area be . Let be the projection of onto the -plane. If we partitioned with rectangles made by and sides (say the area is ), we can think of a rectangular cylinder on a rectangle which cuts a part of . We call this part , and its area .

This rectangular cylinder on the rectangle in the -plane also cuts a part of tangent plane to the surface at a point on . We call this area . If is the angle between the tangent plane and the -plane, then

. (See Figure 1.)

Now is same as the angle between the -axis and a normal line to the surface at the point . But directional cosines of a normal line are proportional to at the point.

Figure 1

Substitute in ,

.

If and are small enough, the area above the tangent plane, , is very close to the surface area . Hence the surface area can be defined by the following limit:

.

Similarly, we can compute area by projecting to other coordinate planes. That is

or

where and are projections of onto the -plane and the -plane respectively.

Find the surface area of a part of circular cone () that lies between and . (See Figure 2.)

Figure 2

Solution. We have .

when , .

From ,

. ■

EXAMPLE2

Find the surface area of the part that was cut by the planes , and in 1st octant. (See Figure 3.)

Figure 3

Solution. From , and

. ■

EXAMPLE3

Find the surface area for the part that was cut by the plane on cylinder in 1st octant.

Figure 4

Solution. The part become a triangle whose boundaries are , and by projecting the part we are looking for onto the -plane. This surface lies above the triangle as

.

For the cylinder surface , we have

.

Hence gives us the following:

. ■

EXAMPLE4

Find the surface area of the sphere .

Solution. Given the equation of sphere , we obtain

.

and so

.

Figure 5

The surface area of in the Figure 5 is one eighth of the total surface area of the sphere with radius .

Using for , the surface area ,

.

Substitute with the polar coordinate system,

. ■

EXAMPLE5

Find a surface area of cylinder inside of . (See Figure 6.)

Figure 6

Solution. Let be the surface area, and be the projection of the part of curve in 1st octant onto the -plane.

Let . Then, .

Using ,

.■

Let be a cylindrical coordinate representation of the equation of a given surface. Then the region can be partitioned with lines that pass through the origin , and concentric circles centered at the origin . Hence we use instead of . In this case the integrand can be represented with and as the followings:

for and ,

.

We have .

In , .

This gives us the following equation.

or

where the region is represented with and as region .

EXAMPLE 6

Find the surface area of the portion of the sphere that is inside the cylinder . (See Figure 7.)

Figure 7

Solution. The equation of a sphere in cylindrical coordinate is .

. Thus

, .

The equation of cylinder in cylindrical coordinates is . Thus,

.

The surface area is

. ■

EXAMPLE7

Let on and be continuous. Find the surface area of the surface of rotation, , which is made by rotating around -axis on . (See Figure 8.)

Figure 8

Solution. The equation of the surface of rotation is

.

Taking partial derivatives of the above equation with respect to and , we have

, .

Then, .

Let and be and to reduce the form of equations, respectively.

Consider the case of ,

.

But

.

Therefore . ■

This result follows from the fact in section 8.2.

14.3 EXERCISES (Surface Area)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-3-Sol.htm

1. Find the surface area of the paraboloid that lies above the -plane.

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-3-1.html

var('x, y')

plot3d(4-x^2-y^2, (x, -2, 2), (y, -2, 2), opacity=0.3, color='yellow')+plot3d(0, (x, -2, 2), (y, -2, 2), opacity=0.3, color='blue')

, , .

.

2. Find the surface area of the circular cylinder that lies under the hemispherical .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-3-2.html

u ,v= var('u, v')

parametric_plot3d( (2*sin(u), 2*cos(u), v), (u, 0, 2*pi), (v, -5, 5), color='red', opacity=0.5)+parametric_plot3d( (v*sin(u), v*cos(u), sqrt(16-v^2)), (u, 0, 2*pi),(v, -3, 3))

,

.

3. Find the area in the first octant among the surface area which was made by intersecting the circular cylinder and the plane .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-3-3.html

var('u, v, x, y')

parametric_plot3d( (sin(u), cos(u), v), (u, 0, 2*pi), (v, -5, 5), color='red', opacity=0.5)+plot3d(2-x-y, (x, -2, 2), (y, -2,2), color='green')

,

.

4. Find the surface area of the portion of the cylinder lying inside the cylinder .

Solution. ,

,

5. Find the surface area of the portion of the cylinder lying inside the sphere .

Solution. ,

,

,

,

6. Find the surface area of a right circular cone of height and base radius .

7. Find the surface area of the upper half of the cone inside the cylinder .

8. Find the surface area of the portion of the sphere that is inside the cylinder .

9. Find the surface area of the portion of the cylinder that is within the cylinder .

10. Find the surface area of the part of the plane that lies in the first octant.

14.4 Cylindrical Coordinates and Spherical Coordinates

Cylindrical and spherical coordinates are two coordinate systems in three dimensions that are similar to polar coordinates in plane. In problems that involve symmetry about an axis, cylindrical coordinates are useful. Normally the -axis is chosen to coincide with this axis of symmetry. In problems where there is symmetry about a point, the spherical coordinate system is useful. For example, the circular cylinder with Cartesian equation has a very simple equation in cylindrical coordinates. (See Figure 1.)

The sphere with center at the origin and radius has the simple equation in spherical coordinates. (See Figure 2.)

Figure 1 , a cylinder. Figure 2

Cylindrical Coordinates

A point in three-dimensional space is represented in the cylindrical coordinate system by the ordered triple , where and are polar coordinates of the projection of onto the -plane and is the directed distance from the -plane to . (See Figure 3.)

Figure 3 The cylindrical coordinates of a point.

The equations

convert cylindrical to rectangular coordinates, while

convert rectangular to cylindrical coordinates.

EXAMPLE1

(a) Find the rectangular coordinates of the point whose cylindrical coordinates are .

(b)Find the cylindrical coordinates of the point whose rectangular coordinates are .

Solution. (a) From , the rectangular coordinates of the given point are

, , .

Thus, the point is in rectangular coordinates.

(b) From , we have

,

, so

.

Therefore, one set of cylindrical coordinates is .

Another is . As with polar coordinates, there are many different expressions. ■

EXAMPLE2

Describe the surface whose equation in cylindrical coordinates is .

Solution. The equation says that the -value, or height, of each point on the surface is the same as , the distance from the point to the -axis. Because doesn’t appear, it can vary. So any horizontal trace in the plane is a circle of radius . These traces suggest that the surface is a cone. This prediction can be confirmed by converting the equation into rectangular coordinates. From the first equation in we have

.

Figure 4 is a cone.

We recognize the equation (by comparison with Table 1 in Section 11.6) as being a circular cone whose axis is the -axis. (See Figure 4.) ■

EXAMPLE3

Find an equation in cylindrical coordinates for the ellipsoid .

Solution. Since from , we have .

So, the equation of the given ellipsoid in cylindrical coordinates is . ■

Spherical Coordinates

In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a point is specified by three numbers. A point in space is represented in spherical coordinates by as shown in Figure 5. Here, is the distance from the origin to is the same angle as in cylindrical coordinates, and is the angle between the positive -axis and the line segment . Note that , .

Figure 5 The spherical coordinates of a point. Figure 6

The spherical coordinates (, , ) of a point can be obtained from its Cartesian coordinates by the formulae , , . From triangles and in Figure 6, we have

, .

But and . Conversely, the Cartesian coordinates may be retrieved from the spherical coordinates (, , ), where ∈ [0, ∞), ∈ , , by

and

The graph of the equation is a vertical half-plane (See Figure 7), and the equation represents a half-cone with the -axis as its axis. (See Figure 8.)

(a) (b)

Figure 7 , a half-plane.

Figure 8 , a half-cone.

EXAMPLE4

Convert the point that is given in spherical coordinates to rectangular coordinates.

Solution. We plot the point in Figure 6. From we have

,

,

.

Figure 9

Thus, the point is in rectangular coordinates. (See Figure 9.) ■

EXAMPLE5

The point is given in rectangular coordinates. Find spherical coordinates for this point.

Solution. From we have and so gives

, ,

, .

(Note that because .) Therefore, spherical coordinates of the given point are . ■

EXAMPLE6

Find an equation in spherical coordinates for the hyperboloid of two sheets with equation .

Solution. Substituting the expressions in into the given equation, we have

or. ■

EXAMPLE7

Find a rectangular equation for the surface with spherical equation given by .

Solution. From and we have

or

which is the equation of a sphere with center and radius . ■

EXAMPLE8

Find the rectangular coordinates of the point in spherical coordinates.

Solution.

T = Spherical('radius', ['azimuth', 'inclination'])

T.transform(radius=1, azimuth=pi/4, inclination=pi/4)

#r, theta=var('r, theta')

#plot3d(r^2, (r, 0, 2), (theta, 0, 2*pi), transformation=T)

Answer : (1/2, 1/2, 1/2*sqrt(2)) ■

EXAMPLE9

Identify the surface whose equation is given by .

http://matrix.skku.ac.kr/cal-lab/cal-14-4-Exm-9.html

Solution.

S=Cylindrical('height', ['radius', 'azimuth'])

r, theta=var('r, theta')

plot3d(r^2, (r, 0, 2), (theta, 0, 2*pi), transformation=S)

Answer : paraboloid ■

Figure 10

14.4 EXERCISES (Cylindrical Coordinates and Spherical Coordinates)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-4-Sol.htm

1-4. Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point.

1. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-1.html

Solution.

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

T = Cylindrical('height', ['radius', 'azimuth'])

T.transform(radius=2, azimuth=- pi/3, height=4)

Answer : The rectangular coordinates of the point of the point is (1, -sqrt(3), 4).

2. .

Solution.

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

T = Cylindrical('height', ['radius', 'azimuth'])

T.transform(radius=1, azimuth= 3*pi/3, height=3)

Answer : .

3. .

Solution. .

4. .

Solution. .

5-6.Change from rectangular to cylindrical coordinates.

5. .

Solution. , so , .

Therefore one set of cylindrical coordinates is . Another is .

As with polar coordinates, there are infinitely many choices.

6. .

7-10. Plot the point whose spherical coordinates are given. Then, find the rectangular coordinates of the point.

7. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-7.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Solution.

T = Spherical('radius', ['azimuth', 'inclination'])

T.transform(radius=3, azimuth=pi/6, inclination=pi/6)

Answer : (3/4*sqrt(3), 3/4, 3/2*sqrt(3))

8. .

Solution. .

9. .

Solution. .

10. .

Solution. .

11-12. Change from rectangular to spherical coordinates.

11. .

Solution. ,

so ,

so (Note that because .)

Therefore, spherical coordinates of the given point are .

12. .

Solution. .

13-14. Change from cylindrical to spherical coordinates.

13. .

Solution. , ,

, .

So, , and .

Therefore, rectangular coordinates is and .

, so .

,

so ( because ).

Therefore, spherical coordinates of the given point are and .

14. .

Solution. .

15-16.Change from spherical to cylindrical coordinates.

15. .

Solution. , ,

, .

So, , since , .

Therefore, rectangular coordinates are

, , so .

Therefore cylindrical coordinates of the given point is .

16. .

Solution. ,

,

Therefore rectangular coordinates of the given point are .

,

, so .

Therefore cylindrical coordinates of the given point is .

17-20. Describe in words the surface whose equation is given.

17. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-17.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Solution.

S=Cylindrical('radius', ['azimuth', 'height'])

theta, z=var('theta, z')

plot3d(4, (theta, 0, 2*pi), (z, -2, 2), transformation=S)

18. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-18.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Solution.

S=Spherical('radius',['azimuth','inclination']);

var('p,theta')

plot3d(4,(p,0,10),(theta,0,2*pi),transformation=S)

19. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-19.html

Solution.

S=Spherical('radius',['azimuth','inclination']);

var('p,theta')

plot3d(4,(p,0,10),(theta,0,2*pi),transformation=S)

20. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-20.html

Solution.

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

S=Spherical('azimuth', ['radius', 'inclination'])

r, phi=var('r, phi')

plot3d(pi/3, (r,0, 10), (phi, 0, pi) , transformation=S)

21-28. Identify the surface whose equation is given.

21. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-21.html

Solution.

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

S=Cylindrical('height', ['radius', 'azimuth'])

r, theta=var('r, theta')

plot3d(2*r^2, (r, 0, 2), (theta,0,2*pi), transformation=S)

22. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-22.html

Solution.

S=Cylindrical('radius', ['azimuth', 'height'])

theta,z=var('theta, z')

plot3d(2*sin(theta), (theta,0,2*pi), (z, -2, 2), transformation=S)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

23. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-23.html

Solution.

S=Spherical('radius', ['azimuth', 'inclination'])

theta, phi=var('theta, phi')

plot3d(1/(2*cos(phi)), (theta, 0, 2*pi), (phi, 0, pi), transformation=S)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

24. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-24.html

Solution.

T = Cylindrical('height',['radius','azimuth']);

var('r,t,z');

implicit_plot3d(r*sin(z)==2,(r,0,3),(t,0,2*pi),(z,-2,2),transformation=T)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

25. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-25.html

Solution.

T=Cylindrical('radius', ['azimuth', 'height'])

theta, z=var('theta, z')

plot3d(3*cos(theta), (theta, 0, 2*pi), (z, -2, 2), transformation=T)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

26. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-26.html

Solution.

T = Cylindrical('height',['radius','azimuth']);

var('r,t,z');

implicit_plot3d(r==3*cos(z),(r,0,3),(t,0,2*pi),(z,-2,2),transformation=T)

27. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-27.html

Solution.

T=Cylindrical('height', ['radius', 'azimuth'])

r, theta=var('r, theta')

p1=plot3d((9-r^2)^(1/2), (r, -3, 3), (theta, 0, 2*pi), transformation=T)

p2=plot3d(-(9-r^2)^(1/2), (r, -3, 3), (theta, 0, 2*pi), transformation=T)

show(p1+p2, aspect_ratio=1)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

28. .

http://matrix.skku.ac.kr/cal-lab/cal-14-2-28.html

Solution.

T = Cylindrical('height',['radius','azimuth']);

var('r,t,z');

implicit_plot3d(r^2*(sin(z)^2+4*cos(z)^2)==2,(r,0,3),(t,0,2*pi),(z,-2,2),transformation=T)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

14.5 Triple Integrals

Similar to the double integral in Section 14.1, we can define triple integrals of a function .

Given a (bounded) function , where is a rectangular parallelepiped in , given by . To define a Riemann sum, we divide into subrectangular parallelepipeds with

, ,

, and , .

Thus we divide a rectangular parallelepiped into sub-parallelepipeds and form the Riemann sum , where is any random point and is the volume of . Then we define the triple integral as the limit of if it exists.

https://www.whitman.edu/mathematics/multivariable/multivariable_15_Multiple_Integration.pdf

DEFINITION1

Let be bounded real-valued functions on a rectangular parallelepiped . If the Riemann defined above converges to a limit , independent of the choice of the sample point , then we say that is integrable over and we write as

or

As before, one can extend the above definition to the more general bounded "elementary domain". If is a bounded elementary domain, then we can choose a rectangular parallelepiped containing . We can extend the integrand function defined on to a function which vanishes outside . Now we can define the triple integral over the general domain as before.

.

These triple integrals can be evaluated as three-fold iterated integrals.

One can also extend Fubini's Theorem for triple integrals. The next example shows how to use Fubini's Theorem to compute triple integrals.

EXAMPLE1

Find .

Solution.

. ■

Note that, if , then the volume of the solid is

.

EXAMPLE2

Find the volume of the region bound by the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through , and .

Solution. The tetrahedron and its projection onto the -plane. The lower boundary is the plane and upper boundary is the plane . Therefore

. ■

EXAMPLE3

Evaluate where is the solid defined by

.

Solution. By using Fubini's Theorem one can the above triple integral as

. ■

EXAMPLE4

Evaluate , where , is the closed region bounded by the cylinder and the planes , , and . (See Figure 1.)

Figure 1

Solution. This closed region is covered if and vary on covering the area and varies from 0 to 2. Therefore

.

http://sagenb.mc.edu/home/pub/36/

http://sage.ace.fordham.edu/home/pub/41/

Answer : 80/3 ■

14.5 EXERCISES (Triple Integrals)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-5-Sol.htm

1. Find .

Solution. http://matrix.skku.ac.kr/cal-lab/cal-14-5-1(new).html

var('x,y,z')

f(x,y)=integral(x*y*z,(z,0,2-x))

g(x)=integral(f(x,y),(y,0,1-x))

integral(g,(x,0,1))

Answer : 13/240

.

2. Find .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-5-2(new).html

var('x,y,z')

f(x,y)=integral(exp(x)*(y+2*z),(z,0,x+y))

g(x)=integral(f(x,y),(y,0,x))

integral(g,(x,0,2))

Answer : 19/3*e^2 + 19

.

3. Find , where .

Solution. http://matrix.skku.ac.kr/cal-lab/cal-14-5-3(new).html

var('x,y,z')

f(x,y)=integral(x+1,(z,-y^2,x^2))

g(x)=integral(f(x,y),(y,0,x))

integral(g,(x,0,1))

Answer : 3/5

.

4. Find , where

.

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-5-4(new).html

var('x,y,z')

f(x,y)=integral(z*x*sin(x*y),(z,0,2))

g(x)=integral(f(x,y),(y,0,pi))

integral(g,(x,1/6,1)).expand()

Answer : 1/pi + 5/3

.

5. Evaluate , where is the solid bounded by parabolic cylinder and the planes .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-5-5(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x,y,z')

assume(x<=2)

assume(x>=0)

f(x,y)=integral(1,(z,0,x))

g(y)=integral(f(x,y),y)

h(x)=g(sqrt(4-2*x))-g(-sqrt(4-2*x))

integral(h,(x,0,2)).expand()

Answer : 64/15

.

6. Find the volume within the ellipsoid

.

Solution.

.

7. Let be a continuous and positive function on and be the region that lies between and -axis.

Find the volume of the solid obtained by rotating about -axis.

Solution.

.

8. Use 7 to find volume of sphere and volume of cylinder , .

9. Find the volume of solid in the first octant () enclosed by a curved surface and the plane .

Solution.

.

10. Find the volume of the solid bounded by the cylinder and planes and .

Solution.

.

11. Find the average value of throughout cubical region bounded by the planes , and in the first octant.

Solution. The volume of the region is . The value of the integral of over the cube is

.

The average value is

.

12.Evaluate where and denotes the closed region bounded by the planes , , , .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-5-1.html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

x,y,z=var('x,y,z');

integral(integral(integral(45*x^2*y, z, 0, 8-4*x-2*y), y, 0, 4-2*x), x, 0, 2)

Answer : 128

14.6 Triple Integrals in Cylindrical and Spherical Coordinates

Cylindrical Coordinates

To evaluate the triple integral in Cartesian coordinates, it is often very useful to use cylindrical coordinates.

Recall that cylindrical coordinates of a point are defined by

.

Under this change of coordinates one can show that the area element in the triple integral changes, to . Thus the triple integral in cylindrical coordinates is obtained by replacing and and changing the limits of integration appropriately.

In particular, the triple integral can be written as

.

EXAMPLE 1

Find .

Solution. Since

,

. ■

EXAMPLE 2

Evaluate , where is bounded by the plane , the -plane and the circular cylinder . (See Figure 1.)

Figure 1

Solution.

. ■

Recall that the following property is in section 14.6.

Let for all points in a region . Then the triple integral is exactly volume of as

.

The next example uses this formula.

EXAMPLE 3

Find the volume of the solid bounded by a spherical surface and the cylinder .

Solution. Since the bounded solid is

and by the symmetry of this solid about -plane (See Figure 2),

Figure 2

Put . Then . When , and when , ,

. ■

Spherical Coordinates

Recall that spherical coordinates of a point are defined by

,

where

Under this change of coordinates to cartesian coordinates one can show that the area element changes to .

The triple integral can be written as

.

Again the appropriate change of limits is required.

EXAMPLE 4

Evaluate , where .

Solution. Using the cylindrical coordinates transformation the given integral can be written as

. . ■

EXAMPLE 5

Find the volume of the sphere with radius .

Solution. The sphere can be described in spherical coordinates as.

Then,

. ■

EXAMPLE 6

Find the volume of the solid enclosed by the spherical surface and the cone . (See Figure 3.)

Figure 3

Solution.

. ■

14.6 EXERCISES (Triple Integrals in Cylindrical and Spherical Coordinates)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-6-Sol.htm

1. Find .

Solution.

.

2. Find the volume of the solid which results by cutting the cylinder by the sphere of radius and centered at the origin.

Solution.

as

,

.

3. Find the volume cut from the cone by the sphere .

Solution.

.

4. Find the volume of the solid bounded by the spherical surface and the circular cone .

Solution.

5. Find the volume of the solid that lies above the cone and below the sphere .

Solution.

6. Evaluate where and is the surface whose side are is the cylinder , and bottom is the disk in the plane , and whose top is the part of the plane that lies above .

http://matrix.skku.ac.kr/cal-lab/cal-14-6-2.html

Solution.

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

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

p1 = implicit_plot3d(z==1+x, (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, 0,2), opacity=0.3, 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)

x = cos(t);

y = sin(t);

z = z;

assume (0 <= t <= 2*pi());

assume (0 <= z <= 1+x);

i,j,k=var('i,j,k');

R=matrix(SR, [[i, j, k], [diff(x, t), diff(y, t), diff(z, t)], [diff(x, z), diff(y, z), diff(z, z)]]).determinant();R

i*cos(t) + j*sin(t)

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

vector_R.norm()

sqrt(abs(sin(t))^2 + abs(cos(t))^2)

integral(integral(z^2*1, z, 0, 1+cos(theta)), theta, 0,2*pi())

Answer : 5/3*pi

7.Find the volume of the region common to the intersecting cylinders and .

8. Find the volume of the region bounded below by the paraboloid and above by the plane .

9. Find the volume cut from the sphere by the cylinder .

10. Evaluate , where is the region inside in the first octant.

11. Find the volume bounded above by the sphere and below by the paraboloid .

Answer : If , . If , .

14.7 Change of Variables in Multiple Integrals

Change of Variables

Recall that integration by substitution (change of variables) is given by

where is a one-one onto -function.

The above formula can also be written as

The idea of change of variables (integration by substitution) can be extended to multiple integrals as well.

First we deal with change of variables for double integrals which can then be extended to triple integrals.

Let be an integrable function. The idea is to change variable and in the double integral into another set of variables every and .

Figure 1

More generally, we consider a transformation . We assume that has continuous 1st order derivative (that is, is a -function). We further assume that maps is a bijection, where is a region in -plane and in -plane.

Now the question is: how this change of variables affects the double integral. For this, we must look at what happens to area element in -plane when it transformed to -plane under .

Look at Figure 2, we consider a rectangle in -plane with area and its image in the -plane.

Figure 2

The idea is to approximate the area of in -plane by a parallelogram determined by secant vectors

and .

Using the definition of partial derivative we have

and

We know that the area of a parallelogram determined by vectors and is given by .

Hence the area of the can be approximated by

provided .

It is easy to see that

.

The determinant on the right hand side is called the Jacobian of and is often denoted by .

This leads us to the following change of variable formula

THEOREM 1 Change of Variables

Let given by be a one-one and onto with a continuous first order partial derivatives ( is ) and . Let be a continuous function. Then

Thus change of variables formula says that we change variable and to and using the transformation and replace the area element by .

Double Integral in Polar Coordinates

The double integrals in polar coordinates becomes a special case of change of variables.

Recall that in polar coordinate we have the transformation

and .

The Jacobian of this transformation is

Hence the double integral in polar coordinates become,

where the limits , and , can be obtains from the given domain of and the transformation .

Figure 3 The formula converts to .

Figure 3 shows a conversion of a square in polar coordinate -plane is a quarter circle of in -plane.

EXAMPLE 1

Evaluate , where is the region enclosed by the line , , and , a parallelogram in the -plane.

Figure 4

Solution. The boundary of the region suggests that

, .

Since this transformation takes the boundary lines

, , ,

to curves and curves

, , ,

in the plane. To find the Jacobian of this transformation, we first solve for and and in terms of and . We get

,

from which we obtain

.

Thus,

. ■

It may not be easy to find a formula for change of variables on a given double integral

.

The method of change of variables is divided into three steps.

(1) Define a transformation on the region in -plane that determines the image of . The boundary of region defines the function. It is important that the image should ensure a simple area to find double integral.

(2) From the transformation , find expressions to integrate.

(3) Find Jacobian and do the integral calculation on the region .

EXAMPLE 2

Evaluate , where is the region enclosed by the rhombus with vertices , , and .

Solution. The region is enclosed by the line ,

,

The expressions and in the equation of the

boundary suggest the transformation

, .

This transformation takes the boundary lines

, , ,

to the curves in the -plane,

, , , . (See Figure 5.)

Figure 5

These lines enclose the region . To find the Jacobian of this transformation, we first solve for and and in terms of and .

We get , from which obtain

.

Thus,

. ■

EXAMPLE 3

Find , where, .

Solution. First of all we use the transformation and . This transformation map. in -plane to the ellipse in the -plane. The Jacobian is

Letting , (i.e. using polar coordinates)

. ■

Triple Integrals

The notation of change of variables of double integrals can be extended to triple integrals.

THEOREM 1 Change of Variable in a Triple Integral

Let given by be an one-one and onto function having first order continuous partial derivatives. We further assume that the jacobian . Let be a continuous function. Then

EXAMPLE 4

Derive the formula for triple integration in spherical coordinates use the above result.

Solution. The change of variables is given by

.

The Jacobian is

Therefore

.

. ■

14.7 EXERCISES(Change of Variables in Multiple Integrals)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-14-7-Sol.htm

1. Evaluate , where is the region enclosed by the lines , and .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-7-1(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x,y')

f(x)=integral((2*x^2-x*y-y^2)/9,x)

g(y)=f(0)-f(-4)

integral(g,(y,1,4))

Answer : 104/9

as

2. Evaluate , where is the region enclosed by the rhombus with vertices , , and .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-7-2(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x,y')

f(y)=integral(x*exp(x*y)/2,y)

g(x)=f(1)-f(-1)

integral(g,(x, 1, 3)).expand()

Answer : 1/2*e^(-3) - 1/2*e^(-1) - 1/2*e + 1/2*e^3

The region is enclosed by the line , , , .

Using Changes of Variables, , as

,

Thus,

.

3. Evaluate , where is the region enclosed by the triangle with vertices , and .

Solution.

http://matrix.skku.ac.kr/cal-lab/cal-14-7-3(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x,y')

f(y)=integral(sin(x)*cos(y)/2,y)

g(x)=f(x)-f(-x)

integral(g,(x,0,2*pi)).expand()

Answer : pi

The region is enclosed by the line , and .

Using Changes of Variables, , as ,

Thus,

.

4. Evaluate the integral , where is the trapezoidal region with vertices , , and .

Solution. http://matrix.skku.ac.kr/cal-lab/cal-14-7-4(new).html

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

var('x,y')

f(x)=integral(cos(x/y)/2,x)

g(y)=f(y)-f(-y)

integral(g,(y,1,2)).expand()

Answer : 3/2*sin(1)

, ,

.

5. Evaluate the integral , where is the trapezoidal region with vertices , , and .

Solution. The region is enclosed by the line , , and .

Using Changes of Variables, , as ,

Thus,

.

6. Derive the formula for triple integration in cylinder coordinates.

7. Evaluate , where is the solid enclosed by the ellipsoid , using the transformation , and .

8. Evaluate , where is the region bounded by the graph of .

9. Evaluate , where is the rhombus region with vertices , , and .

10. Evaluate , where is the triangle region with vertices , and .

Calculus

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Copyright @ 2019 SKKU Matrix Lab. All rights reserved.

Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee http://matrix.skku.ac.kr/sglee/ and http://matrix.skku.ac.kr/cal-book/

*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).