Question

Find the area of the part of the paraboloid z = 9 - x^2 - y^2 that lies above the xy-plane.

Answer 1

Parameterize this surface (call it
`S`) by

`\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(9-u^2)\,\vec k`

with
`0\le u\le3` and
`0\le v\le2\pi`. Take the normal vector to
`S` to be

`\vec r_u*\vec r_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k`

Then the area of
`S` is

`\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u*\vec r_v\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^(2\pi)\int_0^3u√(1+4u^2)\,\mathrm du\,\mathrm dv`

`=\displaystyle2\pi\int_0^3u√(1+4u^2)\,\mathrm du=\boxed{\frac{37√(37)-1}6\pi}`