Question

Find the area of the surface. the part of the hyperbolic paraboloid z = y2 − x2 that lies between the cylinders x2 + y2 = 4 and x2 + y2 = 16.g

Answer 1

Call the surface $S$; then the area of $S$ is given by the surface integral

$\displaystyle\iint_S\mathrm d\mathbf S$

Parameterize the surface by

$\mathbf r(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=u^2\cos^2v-u^2\sin^2v=u^2\cos2v\end{cases}$

with $2\le u\le4$ and $0\le v\le2\pi$. The surface element is given by

$\mathrm d\mathbf S=\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$
$\mathrm d\mathbf S=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

So the area is

$\displaystyle\iint_S\mathrm d\mathbf S=\int_{v=0}^{v=2\pi}\int_{u=2}^{u=4}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{w=17}^{w=65}\sqrt w\,\frac{\mathrm dw}8$

where $w=1+4u^2\implies\mathrm dw=8u\,\mathrm du$

$=\displaystyle\frac\pi4\frac23w^{3/2}\bigg|_{w=17}^{w=65}$
$=\dfrac\pi6\left(65^{3/2}-17^{3/2}\right)\approx237.69$