Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x comma y comma z )dS using a parametric description of the surface. f (x comma y comma z )equalsx squared plus y squared​, where S is the hemisphere x squared plus y squared plus z squared equals 36​, for zgreater than or equals0 The value of the surface integral is nothing. ​(Type an exact​ answers, using pi as​ needed.)

Parameterize [tex]S[/tex] by[tex]\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k[/tex]with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\frac\pi2[/tex]. Take a normal vector to [tex]S[/tex],[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k[/tex]which has norm[tex]\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v[/tex]Then the integral of [tex]f(x,y,z)=x^2+y^2[/tex] over [tex]S[/tex] is[tex]\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}[/tex]