What is the volume above the cone $z= \sqrt{x^2+y^2}$ and bounded by the spheres $^2+y^2+^2=1$ and $^2+y^2+^2=4$?

Question

What is the volume above the cone $z= \sqrt{x^2+y^2}$ and bounded by the spheres $^2+y^2+^2=1$ and $^2+y^2+^2=4$?

I tried converting each equation to cylindrical coordinates: $z= $r, $r^2+^2=4$, and $r^2+^2=1$.

I then set the limits (I don't know which one is the upper limit for z).

$r \leq z \leq$ $1 \leq r \leq 2$ $0 \leq \theta \leq 2 \pi$

Please correct me if I am wrong.

Also this question is not a duplicate of another problem as this question involves finding the volume over a cone and between 2 spheres.

Edit:

If I convert to spherical coordinates would it be:

$0 \leq \phi \leq \pi/4$

$1 \leq \rho \leq 2$

$0 \leq \theta \leq 2 \pi$

Answer 1

We first compute the area that the cone is cutting out from the outer sphere of radius $2$. This region is a spherical cap of height $2-\sqrt{2}$; its area is therefore given by $2\pi\cdot 2\cdot(2-\sqrt{2})$, so that it makes out the part $${4\pi(2-\sqrt{2})\over 4\pi\cdot2^2}={2-\sqrt{2}\over4}$$ of the total outer sphere area. It follows that the volume the cone is cutting out from the outer sphere is $${2-\sqrt{2}\over4}\cdot{4\pi\over3}\cdot 2^3\ .$$ The volume cut out from the inner sphere is ${1\over8}$ of this, so that the volume $V$ we have to compute is given by $$V={7\pi(2-\sqrt{2})\over 3}\ .$$

Answer 2

The shell between the two spheres has volume:

$$\frac{4}{3} \pi (2^3 - 1^3) = \frac{28 \pi}{3}$$

The proportion of this that is within the cone is the area of one sphere's cap to that sphere's area, i.e., $$\frac{2 - \sqrt{2}}{4}$$ so the final volume is:

$$\frac{7 \pi(2 - \sqrt{2})}{3}$$

Answer 3

If you use cylindrical coordinates, you'll have to set it up as a sum of two integrals. When $0\le r\le 1/\sqrt2$, you have $\sqrt{1-r^2}\le z\le\sqrt{4-r^2}$. When $1/\sqrt2\le r\le\sqrt2$, you have $r\le z\le\sqrt{4-r^2}$.

It's far easier to set up in spherical coordinates. Try it.