Surface Area, Surface Integral Examples
                                                        Written by Victoria Kala
                                                         vtkala@math.ucsb.edu
                                                SH 6432u Office Hours: R 12:30 − 1:30pm
                                                         Last updated 6/1/2016
                The first example demonstrates how to find the surface area of a given surface. The second example demon-
                strates how to find the surface integral of a given vector field over a surface.
                   1. Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x2 +y2 = 12
                      and above the xy-plane.
                      Solution. We need to evaluate             ZZ
                                                            A= D||ru×rv||dA.
                      We are asked to find the surface area of a portion of the sphere, this is the surface we need to
                      parametrize. The parametrization vector is given by
                                                             r(x,y,z) = hx,y,zi.
                      In spherical coordinates, x = ρcosθsinφ,y = ρsinθsinφ,z = ρcosφ. Plug these into our parametriza-
                      tion vector:
                                                 r(ρ,θ,φ) = hρcosθsinφ,ρsinθsinφ,ρcosφi.
                      Right now our parametrization is a function of three variables, we need to narrow it down to two
                      variables. We are given one more piece of information about the sphere: it has radius 4. This means
                      ρ = 4. Plug this in:
                                                   r(θ,φ) = h4cosθsinφ,4sinθsinφ,4cosφi.
                      Now we need to find the range of θ and φ. Since we are going all the way around the cylinder and
                      sphere, 0 ≤ θ ≤ 2π. To find the range of φ, we need to find where the cylinder and sphere intersect.
                      The equation of the sphere is
                                                              x2 +y2 +z2 = 16.
                                                     2    2
                      The equation of the cylinder is x +y = 12. Plug this into the equation of the sphere:
                                                   12+z2 =16      ⇒ z2=4 ⇒ z=2.
                      In spherical coordinates, z = ρcosφ, but we know that ρ = 4, so z = 4cosφ. Therefore
                                           z = 2   ⇒ 4cosφ=2 ⇒ cosφ=1 ⇒ φ=π,
                                                                                  2             3
                      and we have that 0 ≤ φ ≤ π. Therefore
                                                3
                                                        A=Z 2πZ π/3||r ×r ||dφdθ.
                                                                         θ    φ
                                                              0   0
                      Now find rθ ×rφ:
                            rθ = h−4sinθsinφ,4cosθsinφ,0i
                            rφ = h4cosθcosφ,4sinθcosφ,−4sinφi
                                                                    
                                       i            j          k    
                                                                                   2              2
                       rθ ×rφ = −4sinθsinφ 4cosθsinφ           0     = h−16cosθsin φ,−16sinθsin φ,−16sinφcosφi
                                                                    
                                                                    
                                   4cosθcosφ    4sinθcosφ −4sinφ
                                                                   1
                        Find ||rθ × rφ||:
                                            q               2   2                2   2                   2
                               ||r ×r || =    (−16cosθsin φ) +(−16sinθsin φ) +(−16sinφcosφ) =... = 16sinφ
                                 θ    φ
                        Therefore
                                      Z 2π Z π/3                   Z 2π Z π/3                               
                                                                                                             π/3
                                 A=              ||r ×r ||dφdθ =              16sinφdφdθ = 2π(−16cosφ)          =16π.
                                                   θ     φ                                                  
                                       0    0                        0    0                                  0
                                           RR                                                                               2   2
                     2. Set up the integral   S F·dS where F = yj−zk and S is the surface given by the paraboloid y = x +z
                        between y = 0 and y = 1. Assume S has positive orientation.
                        Solution. We need to evaluate          ZZ            ZZ
                                                                  S F·dS=       DF·ndA
                        where D the range of the parameters in dA.
                        Weneed to parametrize the surface. The parametrization vector is given by
                                                                  r(x,y,z) = hx,y,zi.
                        We need to narrow this down to dependence on two variables. We are given that the surface is
                             2     2
                        y = x +z . Plug this into the paramterization vector:
                                                                               2    2
                                                                r(x,z) = hx,x +z ,zi.
                        Find r ×r :
                              x     z
                                                               rx = h1,2x,0i
                                                               rz = h0,2z,1i
                                                                              
                                                                    i   j   k
                                                                              
                                                                              
                                                         rx ×rz = 1 2x 0 =h2x,−1,2xi.
                                                                              
                                                                              
                                                                     0  2z   1
                        Then
                           ZZ F·ndA=ZZ h0,y,−zi·h2x,−1,2zidA=ZZ (−y−2z2)dA=ZZ (−(x2+z2)−2z2)dA
                              D               D                                 D                     D
                        Note that we have to plug in y into the integral since we will be integrating over x and z.
                        Wecancertainly find bounds for this integral in Cartesian coordinates, but it will be simpler to switch
                        our integral into cylindrical coordinates. In cylindrical coordinates
                                                    x=rcosθ,z =rsinθ,         0 ≤ r ≤ 1,0 ≤ θ ≤ 2π.
                                                            2    2
                        To get the bounds for r, use y = x +z and y = 1. Therefore
                                     ZZ              ZZ        2    2       2       Z 2π Z 1    2     2    2
                                        DF·ndA= D(−(x +z )−2z )dA= 0                      0 (−r −2r sin θ)drdθ.
                                                                         2