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

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

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^3u\sqrt{1+4u^2}\,\mathrm du=\boxed{\frac{37\sqrt{37}-1}6\pi}$