Question

Use cylindrical coordinates. evaluate (x + y + z) dv e , where e is the solid in the first octant that lies under the paraboloid z = 16 â x2 â y2.

Answer 1

$E=\{(x,y,z)~:~x>0,\,y>0,\,0$

$\displaystyle\iiint_E(x+y+z)\,\mathrm dV=\int_{x=0}^{x=4}\int_{y=0}^{y=\sqrt{16-x^2}}\int_{z=0}^{z=16-x^2-y^2}(x+y+z)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to cylindrical coordinates, we take

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}$

with Jacobian

$\dfrac{\partial(x,y,z)}{\partial(r,\theta,\zeta)}=r$

to get the triple integral

$\displaystyle\iiint_E(x+y+z)\,\mathrm dV=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=4}\int_{\zeta=0}^{\zeta=16-r^2}(r\cos\theta+r\sin\theta+\zeta)r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\left(\int_{\theta=0}^{\theta=\pi/2}(\cos\theta+\sin\theta)\,\mathrm d\theta\right)\left(\int_{r=0}^{r=4}(16r^4-r^2)\,\mathrm dr\right)+\frac\pi2\int_{r=0}^{r=4}\int_{\zeta=0}^{\zeta=16-r^4}r\zeta\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle\frac{4096}{15}+\frac\pi4\int_{r=0}^{r=4}r(16-r^2)^2\,\mathrm dr$
$=\displaystyle\frac{4096}{15}-\frac\pi8\int_{s=16}^{s=0}s^2\,\mathrm ds$

where $s=16-r^2$,

$=\displaystyle\frac{4096}{15}+\frac\pi8\int_{s=0}^{s=16}s^2\,\mathrm ds$
$=\displaystyle\frac{4096}{15}+\frac{4096\pi}{24}$
$=\dfrac{512}{15}(8+5\pi)$