Triple Integral With Spherical Coordinates - Napkin Ring? I wish to calculate the integral bellow, where $T$ is the region bounded by $x^2 + y^2 = 1$ and $x^2 + y^2 + z^2 = 4.$ It looks to me that it represents a Napkin ring.
$$\iiint_T\bigl(x^2 + y^2\bigr)\,\text{d}V.$$
The answer is $\dfrac{\bigl(256 - 132\sqrt{3}\,\bigr)\pi}{15}.$
 A: The definition of $T$ is ambiguous (see also Michael Seifert's comment). According to given answer, $T$ should be the union of two spherical caps and a cylinder:
$$T=\{(x,y,z)\in\mathbb{R}^3:x^2 + y^2 \leq  1,\;x^2 + y^2 + z^2 \leq 4\}.$$
Note that we can split the integral in two:
$$\iiint_T\bigl(x^2 + y^2\bigr)\,\text{d}V=2(I_1+I_2)$$
where
$$I_1=\int_{\phi=0}^{\pi/6}\left(\int_{r=0}^2\left(\int_{\theta=0}^{2\pi}(r\sin(\phi))^2\cdot r^2\sin(\phi)\,d\theta\right)dr\right)d\phi$$
and
$$
I_2=\int_{\phi=\pi/6}^{\pi/2}\left(\int_{r=0}^{1/\sin(\phi)}
\left(\int_{\theta=0}^{2\pi}(r\sin(\phi))^2\cdot r^2\sin(\phi)\,d\theta\right)dr\right)d\phi.$$
Can you take it from here? Check the result with WolframAlpha.
P.S. The Napkin ring is given by
$$N:=\{(x,y,z)\in\mathbb{R}^3:x^2 + y^2 \geq  1,\;x^2 + y^2 + z^2 \leq 4\}.$$ 
and 
$$\iiint_N\bigl(x^2 + y^2\bigr)\,\text{d}V=2I_3=\frac{132\sqrt{3}\pi}{15}$$
where
$$
I_3=\int_{\phi=\pi/6}^{\pi/2}\left(\int_{r=1/\sin(\phi)}^{r=2}
\left(\int_{\theta=0}^{2\pi}(r\sin(\phi))^2\cdot r^2\sin(\phi)\,d\theta\right)dr\right)d\phi.$$
A: Looks to me like the region is the interior, and not the napkin ring.
converting to cylindrical
$\displaystyle\int_0^{2\pi}\int_0^1 \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r^3 \ dz\ dr\ d\theta$
$\displaystyle\int_0^{2\pi}\int_0^1 2r^3\sqrt{4-r^2} \ dr\ d\theta\\
r^2=4-u^2 ,\ 2r\ dr = -2u\ du\\
\displaystyle\int_0^{2\pi}\int_\sqrt{3}^{2} 2(4-u^2)u^2 \ du\ d\theta\\
(4\pi)(\frac 43 u^3 - \frac 15 u^5 )|_\sqrt{3}^2\\
(4\pi)(\frac {64}{15} - \frac {33\sqrt {3}}{15})\\
$
