Question

Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hint: convert to polar coordinates after setting up the integral)

Answer 1

Parameterize the part of the paraboloid within the cylinder - I'll call it $S$ - by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$. The region's area is given by the surface integral

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du$

Take $w=1+36u^2$ so that $\mathrm dw=72u\,\mathrm du$, and the integral becomes

$=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw$
$=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}$
$=\dfrac\pi{54}(145^{3/2}-1)\approx101.522$