SKKU-Calculus-Sec-15-7 Surface Integrals and Flux, SGLee+ 이원준

15.7    Surface Integrals and Flux   by SGLee, 이원준

(Ref. http://math1.skku.ac.kr/home/pub/1884/ 2013.12.11)

## Surface integrals of scalar fields

To find an explicit formula for the surface integral, we need to parameterize the surface of interest, S, by considering a system ofcurvilinear coordinates on S, like the latitude and longitude on a sphere. Let such a parameterization be x(st), where (st) varies in some region T in the plane. Then, the surface integral is given by

where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of x(st), and is known as the surface element.

Surface integrals of vector fields

A vector field on a surface

Consider a vector field v on S, that is, for each x in Sv(x) is a vector.

The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. This applies for example in the expression of the electric field at some fixed point due to an electrically charged surface, or the gravity at some fixed point due to a sheet of material.

Alternatively, if we integrate the normal component of the vector field, the result is a scalar. Imagine that we have a fluid flowing through S, such that v(x) determines the velocity of the fluid at x. The flux is defined as the quantity of fluid flowing through S in unit amount of time.

This illustration implies that if the vector field is tangent to S at each point, then the flux is zero, because the fluid just flows inparallel to S, and neither in nor out. This also implies that if v does not just flow along S, that is, if v has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of v with the unit surface normal to S at each point, which will give us a scalar field, and integrate the obtained field as above. We find the formula

The cross product on the right-hand side of this expression is a surface normal determined by the parametrization.

(Ref. http://en.wikipedia.org/wiki/Surface_integral 2013.12.11)

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

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  .

Sol) $X=(2u,u\mathrm{sin}v,u\mathrm{cos}v)$ 라 하고 $u,v$에 대해서 각각 미분한다.

So, $\frac{\partial X}{\partial u}=(2,\mathrm{sin}v,\mathrm{cos}v)$ and $\frac{\partial X}{\partial v}=(0,u\mathrm{cos}v,-u\mathrm{sin}v)$.

=>  = $\iint_A (u\mathrm{sin}v)(u\mathrm{cos}v)\left \| \frac{\partial X}{\partial u}\times \frac{\partial X}{\partial v} \right \| dudv$

=>$\iint_A (u\mathrm{sin}v)(u\mathrm{cos}v)\left \| \frac{\partial X}{\partial u}\times \frac{\partial X}{\partial v} \right \| dudv = \frac{\sqrt{5}\pi}{2}$  .

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

4. Evaluate, where  is the hemisphere .

Sol)  => $X=(3\mathrm{cos}\theta,3\mathrm{sin}\theta,\sqrt{9-r^2})$  ($0\leq r\leq 3, 0\leq \theta \leq 2\pi$)

(*${\color{Red} u=\theta,v=r}$)

위와 같은 방법으로 $u,v$에 대해서 각각 미분한다.

So, $\left \|\frac{\partial X}{\partial u} \times \frac{\partial X}{\partial v} \right \| = \frac{3v}{\sqrt{9-v^2}}$.

=>$\iint_A 3v dudv$.

=>   $\iint_A 3v dudv = 27\pi$

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.

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 .

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 .

We spilt  into five pieces  as shown in the figure.

,

,

,

Therefore .

9. Find the surface integral over the parallelepiped  when .

,

,

,

,

,

.

Therefore .

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

The sphere can be given a parameterization

.

(*${\color{Red} u=\phi, v=\theta}$)

$\mathbf{r}$을 $u,v$로 각각 미분한 식을 구해보자.

이 두 식의 외적값과 $A(\mathbf{r})$을 내적한 값은 다음과 같다.

(* 교재 풀이에 오류가 있는것 같다. 외적한 값이 옳지 않음)

11. Evaluate .

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

1)  : There  is a graph surface of  on .

We have  and .

Hence by  ,

.

2)  : .

3)  :

.