Question

Evaluate the surface integral. ∫∫s (x2 + y2 + z2) ds s is the part of the cylinder x2 + y2 = 16 that lies between the planes z = 0 and z = 5, together with its top and bottom disks.

Answer 1

Decompose the surface into three components, $\mathbf r_1,\mathbf r_2,\mathbf r_3$, corresponding respectively to the cylindrical region and the top and bottom disks:

$\mathbf r_1(u,v)=\begin{cases}x(u,v)=4\cos u\\y(u,v)=4\sin u\\z(u,v)=v\end{cases}$
where $0\le u\le2\pi$ and $0\le v\le5$,

$\mathbf r_2(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=0\end{cases}$
where $0\le u\le4$ and $0\le v\le2\pi$, and

$\mathbf r_3(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=5\end{cases}$
where $0\le u\le4$ and $0\le v\le2\pi$.

For the cylinder, we have

$\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}=\langle4\cos u,4\sin u,0\rangle\implies\left\|\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}\right\|=4$

and the integral over this surface is

$\displaystyle\iint_{\text{cyl}}(x^2+y^2+z^2)\,\mathrm dS=4\int_{v=0}^{v=5}\int_{u=0}^{u=2\pi}((4\cos u)^2+(4\sin u)^2+v^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle320\int_{u=0}^{u=2\pi}\mathrm du+8\pi\int_{v=0}^{v=5}v^2\,\mathrm dv$
$=640\pi+\dfrac83\pi(125)$
$=\dfrac{2920\pi}3$

Bottom disk:

$\dfrac{\partial\mathbf r_2}{\partial u}\times\dfrac{\partial\mathbf r_2}{\partial v}=\langle0,0,u\rangle\implies\left\|\dfrac{\partial\mathbf r_2}{\partial u}\times\dfrac{\partial\mathbf r_2}{\partial v}\right\|=u$

and the integral over the bottom disk is

$\displaystyle\iint_{z=0}(x^2+y^2+z^2)\,\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=4}u((u\cos v)^2+(u\sin v)^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=4}u^3\,\mathrm du$
$=128\pi$

The setup for the integral along the top disk is similar to that for the bottom disk, except that $z=5$:

$\displaystyle\iint_{z=5}(x^2+y^2+z^2)\,\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=4}u((u\cos v)^2+(u\sin v)^2+5^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=4}(u^3+25u)\,\mathrm du$
$=528\pi$

Finally, the value of the integral over the entire surface is the sum of the integrals over the component surfaces:

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\frac{2920\pi}3+128\pi+528\pi=\dfrac{4888\pi}3$