Let $a>R>0$. Calculate the volume of the body that meets the differences $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$

Question

Let $a > R > 0$. Calculate the volume of the body constricted by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$

I don't know where to start with this problem. The constraints form a cylinder which body I need to integrate the volume of. Only hint that I've gotten is that it might be helpful to use polar-coordinates. Would really appreciate help!

Answer 1

Here is an easy way to calculate this volume, IMO.

Let's denote $I$ be this volume , and $J$ be the complement of $I$ (in the sphere). We have then $I+J = \frac{4}{3}\pi a^3$.

The volume $J$ is defined by the 2 inequalities: $x^2+y^2+z^2 \le a^2$ and $x^2+y^2 >R^2$. We deduce then $a^2-z^2 > R^2$ or $ -\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2} $

We have \begin{align} J &= \int_{-\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2}} \left( \int_{R^2<x^2+y^2 \le a^2-z^2}dxdy \right) dz\\ \end{align} It's evident that the inner integral is the surface are of a circle ring and is equal to $\pi \left((a^2-z^2) -R^2\right)$. Then \begin{align} J &= \int_{-\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2}} \pi \left((a^2-R^2) -z^2\right) dz \\ &= 2\pi (a^2-R^2)\sqrt{a^2-R^2}- \frac{2}{3} \pi (a^2-R^2)^{\frac{3}{2}} \\ &= \frac{4}{3} \pi (a^2-R^2)^{\frac{3}{2}} \\ \end{align}

Finally, we have $$I = \frac{4}{3}\pi a^3 - J = \frac{4}{3}\pi a^3 -\frac{4}{3} \pi (a^2-R^2)^{\frac{3}{2}} $$

Answer 2

I'm assuming you mean $x^2+y^2+z^2\leq a$. In this case we are looking for the volume between the cylinder and the sphere. I would use cylindrical coordinates (https://en.wikipedia.org/wiki/Cylindrical_coordinate_system) which are the same thing as polar except there is also a z coordinate. Then $x^2+y^2\leq R$ becomes $r^2\leq R$. Similarly, $x^2+y^2+z^2 \leq a$ becomes $r^2+z^2 \leq a$. Therefore, $z^2 \leq a-r^2$. This gives you an upper and lower bound on $z$. Finally, if you visualize the shape you can figure out bounds for $\theta$. Your integral should be in the form $$\int_{\text{lower theta bound}}^{\text{upper theta bound}} \int_{\text{lower r bound}}^{\text{upper r bound}} \int_{-\sqrt{a-r^2}}^{\sqrt{a-r^2}} 1 dz dr d\theta$$ I'll let you figure out the remaining bounds.

We integrate over $1$ because that gives us volume in the same way that in $\mathbb{R}^2$ integrating over 1 gives the area of the region we are integrating over.

Note that the square roots are real because $r^2\leq R^2 \leq a^2$.