Answer
a) $\frac{512\pi}{3}$
b)Centroid: $(\bar x, \bar y, \bar z)=(0,0,\frac{23}{2})$
Work Step by Step
a)
$z=24-x^{2}-y^{2}$ becomes $z=24-r^{2}$
and $z=2\sqrt {x^{2}+y^{2}}$ becomes $z=2r$
Solve for $r$:
$24-r^{2}=2r$
$r^{2}+2r-24=0$
$(r+6)(r-4)=0$
$r=-6,4$
But $r=-6$ is extraneous. So $r=4$.
The bounds for region E are
$E=\{(r,\theta,z)|\ 0\leq r \leq 4, 0 \leq \theta \leq 2\pi, 2r \leq z \leq 24-r^{2}\}$
The volume of the region E is:
$V(S)=\iiint_E dV=\int_{0}^{2\pi}\int_{0}^{4}\int_{2r}^{24-r^{2}}r\ dz\ dr\ d\theta$
$=\int_{0}^{2\pi}\int_{0}^{4}r(z)|^{24-r^{2}}_{2r}dr\ d\theta$
$=\int_{0}^{2\pi}\int_{0}^{4}r(24-r^2-2r)dr\ d\theta$
$=\int_{0}^{2\pi}\int_{0}^{4}(24r-r^3-2r^2)dr\ d\theta$
$=\int_{0}^{2\pi}(\frac{24r^2}{2}-\frac{r^4}{4}-\frac{2r^3}{3})|^4_0\ d\theta$
$=\int_{0}^{2\pi}(12r^2-\frac{r^4}{4}-\frac{2r^3}{3})|^4_0\ d\theta$
$=\int_{0}^{2\pi}(12\times 4^2-\frac{4^4}{4}-\frac{2\times4^3}{3})\ d\theta$
$=(12\times 4^2-\frac{4^4}{4}-\frac{2\times4^3}{3})\int_{0}^{2\pi}d\theta$
$=(3\times4\times 4^2-4^3-\frac{2\times4^3}{3})(\theta)|^{2\pi}_{0}$
$=2\pi(3\times 4^3-4^3-\frac{2\times4^3}{3})$
$=2\pi(2\times 4^3-\frac{2\times4^3}{3})$
$=2\pi(\frac{6\times4^3}{3}-\frac{2\times4^3}{3})$
$=2\pi(\frac{4\times4^3}{3})$
$=2\pi\times\frac{4\times64}{3}$
$=2\pi\times\frac{256}{3}$
$=\frac{512\pi}{3}$
b)
$(\bar x, \bar y, \bar z) = (\frac{\iiint_E x\ dV}{\iiint_E dV},\frac{\iiint_E y\ dV}{\iiint_E dV},\frac{\iiint_E z\ dV}{\iiint_E dV})$
$= (\frac{\int_{0}^{2\pi}\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr\ d\theta}{2\pi(\frac{4\times4^3}{3})},
\frac{\int_{0}^{2\pi}\int_{0}^{4}\int_{2r}^{24-r^{2}}r(sin\theta)\times r\ dz\ dr\ d\theta}{2\pi(\frac{4\times4^3}{3})},
\frac{\int_{0}^{2\pi}\int_{0}^{4}\int_{2r}^{24-r^{2}}z\times r\ dz\ dr\ d\theta}{2\pi(\frac{4\times4^3}{3})})$
$= (\frac{\int_{0}^{2\pi}cos\theta\ d\theta\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{\int_{0}^{2\pi}sin\theta\ d\theta\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{\int_{0}^{2\pi}d\theta\times\int_{0}^{4}\int_{2r}^{24-r^{2}}z\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})})$
$= (\frac{(cos\theta)|^{2\pi}_{0}\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{(sin\theta)|^{2\pi}_{0}\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{2\pi\times\int_{0}^{4}(r\int_{2r}^{24-r^{2}}z\ dz)\ dr}{2\pi(\frac{4\times4^3}{3})})$
$= (\frac{(cos(2\pi)-cos0)\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{(sin(2\pi)-sin0)\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},
\frac{\int_{0}^{4}(r(z^{2}/2)|_{2r}^{24-r^{2}})\ dr}{\frac{4\times4^3}{3}})$
$= (\frac{(1-1)\times\int_{0}^{4}\int_{2r}^{24-r^{2}}r(cos\theta)\times r\ dz\ dr}{2\pi(\frac{4\times4^3}{3})},\ 0,
\frac{\int_{0}^{4}(\frac{r}{2}(z^{2})|_{2r}^{24-r^{2}})\ dr}{\frac{4\times4^3}{3}})$
$= (0,\ 0,\frac{\int_{0}^{4}(r((24-r^2)^2-(2r)^2))\ dr}{2\times\frac{4\times4^3}{3}})$
$= (0,\ 0,\frac{\int_{0}^{4}r(24-r^2)^2\ dr\ -\int_{0}^{4}r(2r)^2\ dr}{2\times\frac{4\times4^3}{3}})$
$= (0,\ 0,\frac{\int_{0}^{4}r(24-r^2)^2\ dr\ -\int_{0}^{4}r(4r^2)\ dr}{2\times\frac{4\times4^3}{3}})$
$= (0,\ 0,\frac{\int_{0}^{4}r(24-r^2)^2\ dr\ -\int_{0}^{4}r(4r^2)\ dr}{2\times\frac{4\times4^3}{3}})$
Perform u-substitution:
$u=24-r^2$
$du=-2r\ dr$
$\frac{du}{-2}=r\ dr$
$u(4)=24-4^2=24-16=8$
$u(0)=24-0^2=24$
$(0,\ 0,\frac{\int_{0}^{4}r(24-r^2)^2\ dr\ -\int_{0}^{4}r(4r^2)\ dr}{2\times\frac{4\times4^3}{3}})$
$=(0,\ 0,\frac{\int_{24}^{8}u^2\times\frac{du}{-2}\ -\int_{0}^{4}(4r^3)\ dr}{2\times\frac{4\times4^3}{3}})$
$=(0,\ 0,\frac{\frac{1}{-2}\int_{24}^{8}u^2\ du\ -(r^4)|_{0}^{4}}{2\times\frac{4\times4^3}{3}})$
$=(0,\ 0,\frac{\frac{1}{-2}(\frac{u^3}{3})|_{24}^{8}\ -4^4}{2\times\frac{4\times4^3}{3}})$
$=(0,\ 0,\frac{\frac{1}{-2\times3}(u^3)|_{24}^{8}\ -4^4}{2\times\frac{4\times4^3}{3}})$
$=(0,\ 0,\frac{\frac{1}{-2\times3}(8^3-24^3)\ -4^4}{2\times\frac{4^4}{3}})$
$=(0,\ 0,\frac{\frac{1}{-2\times3}(4^3\times2^3-4^3\times2^3\times3^3)}{2\times\frac{4^4}{3}}-\frac{4^4}{2\times4^4/3})$
$=(0,\ 0,\frac{4^3\times2^3(1-3^3)}{-2\times3\times2\times\frac{4^4}{3}}-\frac{1}{2/3})$
$=(0,\ 0,\frac{2(1-3^3)}{-4}-\frac{3}{2})$
$=(0,\ 0,\frac{(3^3-1)}{2}-\frac{3}{2})$
$=(0,\ 0,\frac{(27-1)}{2}-\frac{3}{2})$
$=(0,\ 0,\frac{23}{2})$
