Lecture 22, November 23
                                                                                    3
                 • Surface integrals. The integral of f(x,y,z) over a surface σ in R is
                             ∫∫ f(x,y,z)dS = ∫∫ f(x(u,v),y(u,v),z(u,v))·||r ×r ||dudv,
                                                                                u    v
                                σ
                    where r(u,v) = ⟨x(u,v),y(u,v),z(u,v)⟩ is the parametric equation of the surface.
                                                                             √       2    2
                 • When the surface is the graph of z = f(x,y), one has dS =    1+fx +fy dxdy.
                    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
                                                                                      2    2
                    Example 1 (Cylinder). The parametric equation of the cylinder x +y = 1 is
                                               r = ⟨x,y,z⟩ = ⟨cosθ,sinθ,z⟩
                    and it is obtained using cylindrical coordinates. In this case, we have
                                    r ×r =⟨−sinθ,cosθ,0⟩×⟨0,0,1⟩=⟨cosθ,sinθ,0⟩,
                                     θ    z   √
                                                   2       2
                                 ||r ×r || =    cos θ +sin θ = 1
                                    θ    z
                    and one can use these facts to compute any surface integral over the cylinder.
                                                                                 √ 2      2
                    Example 2 (Cone). The parametric equation of the cone z =       x +y is
                                  x2 +y2 = z2    =⇒ r=⟨x,y,z⟩=⟨zcosθ,zsinθ,z⟩.
                    To compute a surface integral over the cone, one needs to compute
                             r ×r =⟨−zsinθ,zcosθ,0⟩×⟨cosθ,sinθ,1⟩ = ⟨zcosθ,zsinθ,−z⟩,
                              θ    z   √
                                          2   2      2   2     2    √
                          ||r ×r || =    z cos θ +z sin θ +z = z 2.
                             θ    z
                                                                                  2    2    2
                    Example 3 (Sphere). The parametric equation of the sphere x +y +z = 1 is
                                        r = ⟨x,y,z⟩ = ⟨cosθsinϕ,sinθsinϕ,cosϕ⟩
                    and it is obtained using spherical coordinates. In this case, we have
                            r ×r =⟨−sinθsinϕ,cosθsinϕ,0⟩×⟨cosθcosϕ,sinθcosϕ,−sinϕ⟩
                             θ    ϕ   ⟨                                     ⟩
                                                2            2
                                   = −cosθsin ϕ,−sinθsin ϕ,−sinϕcosϕ =−(sinϕ)r
                    and the fact that ||r|| = 1 implies that ||r × r || = sinϕ.
                                                            θ    ϕ
                     Example 4 (A general example). The graph of z = f(x,y) can be described by
                                                  r = ⟨x,y,z⟩ = ⟨x,y,f(x,y)⟩.
                     To compute a surface integral over this graph, one needs to compute
                                          r ×r =⟨1,0,f ⟩×⟨0,1,f ⟩ = ⟨−f ,−f ,1⟩,
                                           x    y    √ x               y        x    y
                                       ||r ×r || =     1+f2+f2.
                                          x    y            x    y
                     Example 5. We compute the integral ∫∫σz2dS in the case that σ is the part of the
                              √ 2      2
                     cone z =    x +y that lies between z = 0 and z = 1. As in Example 2, we have
                                                                                    √
                                          r = ⟨zcosθ,zsinθ,z⟩,        ||r ×r || = z 2.
                                                                        θ    z
                     This implies dS = z√2dzdθ, so the given integral becomes
                                     ∫∫           ∫ 2π ∫ 1  √           ∫ 2π √          √
                                         z2dS =           z3  2dzdθ =          2 dθ = π 2.
                                        σ          0    0                 0   4          2
                     Example 6. Consider the lamina that occupies the part of the paraboloid z = x2+y2
                     that lies below the plane z = 1. If its density is given by δ(x,y,z), then its mass is
                                                    Mass = ∫∫ δ(x,y,z)dS.
                                                               σ
                                                                                     2    2
                     Assume that δ is constant for simplicity. Since z = f(x,y) = x + y , we have
                                                       √      2     2   √        2     2
                                         ||r ×r || =     1+f +f = 1+4x +4y
                                            x    y            x    y
                     by Example 4. Using this fact and switching to polar coordinates, we find that
                                               ∫ ∫ √           2     2
                                       Mass =        δ   1+4x +4y dxdy
                                                 ∫ 2π ∫ 1       2 1/2
                                             =δ          (1 +4r )    rdrdθ
                                                  0    0
                                                δ ∫ 2π ∫ 5 1/2                           2
                                             =8           u   dudθ            u=1+4r
                                                   0    1
                                                 ∫ 2π 3/2     3/2
                                             = δ      5    −1 dθ
                                                8 0       3/2
                                                δπ   √
                                             = 6 (5 5−1).