Question

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

Answer 1

You can parameterize
`S` using spherical coordinates by

`\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle`

with
`0\le u\le2\pi` and
`0\le v\le\frac\pi2`.

Take the normal vector to
`S` to be

`(\partial\vec s)/(\partial\vec v)*(\partial\vec s)/(\partial\vec u)=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle`

(I use
`\vec s_v*\vec s_u` to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use
`\vec s_u*\vec s_v=-(\vec s_v*\vec s_u)`.)

Then

`f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v`

and the integral of
`f` over
`S` is

`\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^(\pi/2)\int_0^(2\pi)36\sin^2v\left\|(\partial\vec s)/(\partial v)*(\partial\vec s)/(\partial u)\right\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^(\pi/2)\int_0^(2\pi)(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv`

`=\displaystyle2592\pi\int_0^(\pi/2)\sin^3v\,\mathrm dv=\boxed{1728\pi}`