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)
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(s, t), where (s, t) 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(s, t), and is known as the surface element.
Surface integrals of vector fields
Consider a vector field v on S, that is, for each x in S, v(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
2. Evaluate , where is the surface with parametric equations , and , .
Sol) 라 하고 에 대해서 각각 미분한다.
So, and .
3. Evaluate, where is the part of the plane that lies in the first octant.
4. Evaluate, where is the hemisphere .
Sol) => ()
위와 같은 방법으로 에 대해서 각각 미분한다.
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 .
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.
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.
9. Find the surface integral over the parallelepiped , , , , , when .
10. If is the surface of the sphere and , evaluate .
The sphere can be given a parameterization
을 로 각각 미분한 식을 구해보자.
이 두 식의 외적값과 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) : .