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*\mathbf s_v\|=u√(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√(1+4u^2)\,\mathrm du\,\mathrm dv`

`=\displaystyle2\pi\int_(u=0)^(u=2)u√(1+4u^2)\,\mathrm du=\frac{(17^(3/2)-1)\pi}6\approx36.1769`