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

 \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. 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)  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)  : 

                            .




Back to Part II