find the volume of the solid of intersection of the two spheres of radii a and b (with b $<$ a)

Question

Also the center of the smaller one lies on the surface of the larger one.

This is what I have thus far.

Utilize the Cartesian plane we have for the smaller and larger spheres respectively: $(x-a)^2 + y^2 = b^2 $ and $x^2 + y^2 = a^2$

I solve for the $x$ value where these 2 intersect. I get $x_r = \frac{2a^2 - b^2}{2a}$

To get the volume of the larger cap in the intersection I used the method of cross sections. $\int^{a}_{x_r} \pi(a^2 - x^2)dx$

Similarly for the smaller sphere: $\int^{b}_{x_r} \pi(b^2 - (x-a)^2)dx$

Afterwards, I sum the result of the two to get the answer? Did I make a mistake somewhere

Answer 1

Your approach is fine. You are basically taking $x$ as your $z$ axis and hence in cylindrical coordinates $y^2 + z^2 = r^2$ and so you may want to write them as,

$r^2+x^2 = a^2$ and $r^2 + (x-a)^2 = b^2$.

Your first integral is fine but your second integral should be

$\displaystyle \int_{a-b}^{\frac{2a^2 - b^2}{2a}} \pi \, \big(b^2 - (a-x)^2\big) \,dx = \frac{\pi b^6}{24a^3} - \frac{\pi b^4}{2a} + \frac{2 \pi b^3}{3}$

The lower bound is the minimum value of $x$ for the smaller sphere, which is $(a-b)$ and it is part of the intersection volume.

Answer 2

If the circles of radii $b,\,a$ have respective centres $(a,\,0,\,0),\,(0,\,0,\,0)$, their intersection is confined to $|z|\le b<a$. For such $z$, the intersection satisfies$$0\le(x-a)^2+y^2\le b^2-z^2,\,0\le x^2+y^2\le a^2-z^2.$$Rewriting the $xy$-plane with polar coordinates, the first double inequality is$$\begin{align}(r-a\cos\theta)^2&\le b^2-z^2-a^2\sin^2\theta\\\iff r&\in\left[a\cos\theta-\sqrt{b^2-z^2-a^2\sin^2\theta},\,a\cos\theta+\sqrt{b^2-z^2-a^2\sin^2\theta}\right],\end{align}$$which is satisfiable iff $\sin^2\theta\le\frac{b^2-z^2}{a^2}$. The second sphere's double inequality is equivalent to $0\le r\le\sqrt{a^2-z^2}$. So the intersection is$$\begin{align}|\theta|&\le\arcsin\frac{\sqrt{b^2-z^2}}{a},\\r&\in\left[\max\left\{0,\,a\cos\theta-\sqrt{b^2-z^2-a^2\sin^2\theta}\right\},\,\min\left\{\sqrt{a^2-z^2},\,a\cos\theta+\sqrt{b^2-z^2-a^2\sin^2\theta}\right\}\right].\end{align}$$If we right this more concisely as $-f(z)\le\theta\le f(z),\,g_-(\theta)\le r\le g_+(\theta)$, the volume of the intersection is$$\int_{-b}^bdz\int_{-f}^fd\theta\int_{-g}^grdr.$$