Question

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.

Answer 1

I assume the cone has equation
`z=√(x^2+y^2)` (i.e. the upper half of the infinite cone given by
`z^2=x^2+y^2`). Take

`\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`

The volume of the described region (call it
`R`) is

`\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`

The limits on
`\theta` and
`\rho` should be obvious. The lower limit on
`\varphi` is obtained by first determining the intersection of the cone and sphere lies in the cylinder
`x^2+y^2=1`. The distance between the central axis of the cone and this intersection is 1. The sphere has radius
`\sqrt2`. Then
`\varphi` satisfies

`\sin\varphi=\frac1{\sqrt2}\implies\varphi=\frac\pi4`

(I've added a picture to better demonstrate this)

Computing the integral is trivial. We have

`\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}`