Use spherical coordinates to find the volume of the region that lies outside the cone z = p x 2 + y 2 but inside the sphere x 2 + y 2 + z 2 = 2. Write the answer as an exact answer, which should involve π and √ 2. Do not round or use a calculator.

I assume the cone has equation [tex]z=\sqrt{x^2+y^2}[/tex] (i.e. the upper half of the infinite cone given by [tex]z^2=x^2+y^2[/tex]). Take[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]The volume of the described region (call it [tex]R[/tex]) is[tex]\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta[/tex]The limits on [tex]\theta[/tex] and [tex]\rho[/tex] should be obvious. The lower limit on [tex]\varphi[/tex] is obtained by first determining the intersection of the cone and sphere lies in the cylinder [tex]x^2+y^2=1[/tex]. The distance between the central axis of the cone and this intersection is 1. The sphere has radius [tex]\sqrt2[/tex]. Then [tex]\varphi[/tex] satisfies[tex]\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4[/tex](I've added a picture to better demonstrate this)Computing the integral is trivial. We have[tex]\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}[/tex]