Question

Find the area of the surface. the part of the paraboloid z=4-x^2-y^2 that lies above the xy-plane

Answer 1

Parameterize the surface (call it $\mathcal S$) by

$\mathbf s(u,v)=\langle u\cos v,u\sin v,4-u^2\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$. Then the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The area of $\mathcal S$ is then given by the surface integral

$\displaystyle\iint_{\mathcal S}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du=\dfrac{(17^{3/2}-1)\pi}6\approx36.1769$