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*\mathbf r_v\right\|\,\mathrm dv\,\mathrm du`

`=\displaystyle\int_(u=0)^(u=1)\int_(v=0)^(v=2\pi)u√(1+4u^2)\,\mathrm dv\,\mathrm du`

`=\displaystyle2\pi\int_(u=0)^(u=1)u√(1+4u^2)\,\mathrm du`

`=\frac{(5\sqrt5-1)\pi}6`