Question

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.)

Answer 1

Parameterize
`S` by

`\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k`

with
`0\le u\le2\pi` and
`0\le v\le\frac\pi2`. Take a normal vector to
`S`,

`(\partial\vec s)/(\partial v)*(\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`

which has norm

`\left\|(\partial\vec s)/(\partial v)*(\partial\vec s)/(\partial u)\right\|=36\sin v`

Then the integral of
`f(x,y,z)=x^2+y^2` over
`S` is

`\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\|(\partial\vec s)/(\partial v)*(\partial\vec s)/(\partial u)\right\|\,\mathrm du\,\mathrm dv`

`=\displaystyle36^2\int_0^(\pi/2)\int_0^(2\pi)\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}`