Question

Find the surface area of the part of the circular paraboloid z=x2+y2 that lies inside the cylinder x2+y2=1.

Answer 1

Parameterize the region by

$\mathbf r(u,v)=(x(u,v),y(u,v),z(u,v))=(u\cos v,u\sin v,u^2)$

where $0\le u\le1$ and $0\le v\le2\pi$. The area of the surface is then given by the surface integral

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm dv\,\mathrm du$
$=\displaystyle\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}u\sqrt{1+4u^2}\,\mathrm dv\,\mathrm du$
$=\displaystyle2\pi\int_{u=0}^{u=1}u\sqrt{1+4u^2}\,\mathrm du$
$=\dfrac{(5\sqrt5-1)\pi}6$