Triple Integral With Spherical Coordinates - Napkin Ring?

Question

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

Answer 1

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 $$ I_3=\int_{\phi=\pi/6}^{\pi/2}\left(\int_{r=1/\sin(\phi)}^{r=2} \left(\int_{\theta=0}^{2\pi}(r\sin(\phi))^2\cdot r^2\sin(\phi)\,d\theta\right)dr\right)d\phi.$$

Answer 2

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})\\ $