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

  15.7    Surface Integrals and Flux   by SGLee, 이원준

(Ref. 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

 \iint_{S} f \,dS = \iint_{T} f(\mathbf{x}(s, t)) \left\|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right\| ds\, dt

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

\iint_S {\mathbf v}\cdot \,d{\mathbf {S}} = \int_S {\mathbf v}\cdot {\mathbf n}\,dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over
\partial s}\times {\partial \mathbf{x} \over \partial t}\right) ds\, dt.

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

(Ref. 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.

So, .



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







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)  : .


 3)  : 


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