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 $\dfrac{\bigl(256 - 132\sqrt{3}\,\bigr)\pi}{15}.$

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})\\
$
