Question

Find the area of the surface. the part of the plane 4x + 4y + z = 16 that lies inside the cylinder x2 + y2 = 9

Answer 1

As a first step, convert to cylindrical coordinates. The equation $x^2+y^2=9=3^2$ is a clue to set

$x=3r\cos\theta$

$y=3r\sin\theta$

then have $0\le r\le1$ and $0\le\theta\le2\pi$. Meanwhile, the plane equation tells you to use

$4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta$

So we parameterize the part of the plane within the cylinder with the vector-valued function

$\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)$

The surface integral giving the area of the surface is


$\displaystyle\iint_{\mathcal S}\mathrm dS$

where $\mathcal S$ denotes the surface in question, and the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right\|\,\mathrm dr\,\mathrm d\theta$

We have

$\dfrac{\partial\mathbf s}{\partial r}=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))$

$\dfrac{\partial\mathbf s}{\partial\theta}=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))$

$\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9\sqrt{33}r\,\mathrm dr\,\mathrm d\theta$

So the area is

$\displaystyle\iint_{\mathcal S}\mathrm dS=9\sqrt{33}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta=18\sqrt{33}\,\pi$