Question

Evaluate the surface integral. (give your answer correct to at least three decimal places.) s is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 2

Answer 1

Split up the surface $S$ into three main components $S_1,S_2,S_3$, where

$S_1$ is the region in the plane $y=0$ bounded by $x^2+z^2=1$;

$S_2$ is the piece of the cylinder bounded between the two planes $y=0$ and $x+y=2$;

and $S_3$ is the part of the plane $x+y=2$ bounded by the cylinder $x^2+z^2=1$.

These surfaces can be parameterized respectively by

$S_1:~\mathbf s_1(u,v)=\langle u\cos v,0,u\sin v\rangle$
where $0\le u\le1$ and $0\le v\le2\pi$,

$S_2:~\mathbf s_2(u,v)=\langle\cos v,u,\sin v\rangle$
where $0\le u\le2-\cos v$ and $0\le v\le2\pi$,

$S_3:~\mathbf s_3(u,v)=\langle u\cos v,2-u\cos v,u\sin v\rangle$
where $0\le u\le1$ and $0\le v\le2\pi$.

The surface integral of a function $f(x,y,z)$ along a surface $R$ parameterized by $\mathbf r(u,v)$ is given to be

$\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\iint_Sf(\mathbf r(u,v))\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times\frac{\partial\mathbf r(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

Assuming we're just finding the area of the total surface $S$, we take $f(x,y,z)=1$, and split up the total surface integral into integrals along each component surface. We have

$\displaystyle\iint_{S_1}\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}u\,\mathrm dv\,\mathrm du$
$\displaystyle\iint_{S_1}\mathrm dS=\pi$

$\displaystyle\iint_{S_2}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2-u\cos v}\mathrm du\,\mathrm dv$
$\displaystyle\iint_{S_2}\mathrm dS=4\pi$

$\displaystyle\iint_{S_3}\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\sqrt2u\,\mathrm dv\,\mathrm du$
$\displaystyle\iint_{S_3}\mathrm dS=\sqrt2\pi$

Therefore

$\displaystyle\iint_S\mathrm dS=\left\{\iint_{S_1}+\iint_{S_2}+\iint_{S_3}\right\}\mathrm dS=(5+\sqrt2)\pi\approx20.151$