Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric description of the surface.f(x,y,z)=x2+y2, where S is the hemisphere x2+y2+z2=36, for z>=0

You can parameterize [tex]S[/tex] using spherical coordinates by[tex]\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle[/tex]with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\frac\pi2[/tex].Take the normal vector to [tex]S[/tex] to be[tex]\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle[/tex](I use [tex]\vec s_v\times\vec s_u[/tex] to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use [tex]\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)[/tex].)Then[tex]f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v[/tex]and the integral of [tex]f[/tex] over [tex]S[/tex] is[tex]\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle2592\pi\int_0^{\pi/2}\sin^3v\,\mathrm dv=\boxed{1728\pi}[/tex]