Is there a way to calculate the enclosed volume of two unit balls centered at $(0,0,0)$ and $(1,1,1)$ using a triple integral?

Question

Let $\Omega \subset \mathbb{R}^3$ be the subset $$\Omega = B((0,0,0),1) \cap B((1,1,1),1)$$ where $B(x,r)$ stands for the ball centered in $x$ with radius $r$.

I was looking for an easy way to calculate the volume with a triple integral.

Also because the exercise asks, after this, to calculate $$\int_{\Omega} (x+2y+z) dV$$ and $$\int_{\partial \Omega} (x+2y+z) d\Sigma$$

Since this was an exercise from a test that does not last long, there must be some shortcut that I'm missing here.

Does someone have some ideas? Thanks!

Answer 1

Due to the perfect symmetry of the two-ball configuration around the axis (1,1,1), these three integrals can be carried out relatively straightforwardly.

Such symmetry is obvious for the simple volume integral,

$$I_1=\int_{\Omega} dV$$

For the other two integrals, we observe that

$$\int_{\Omega} xdV = \int_{\Omega} ydV = \int_{\Omega} zdV$$

and rewrite them in symmetric forms explicitly

$$I_2=\int_{\Omega} (x+2y+z) dV=\frac{4}{3}\int_{\Omega} (x+y+z) dV$$

$$I_3=\int_{\partial \Omega} (x+2y+z) d\Sigma=\frac{4}{3}\int_{\partial\Omega} (x+y+z) d\Sigma$$

Observe further that $x+y+z = a$ is a plane intersecting all three axes at $a$ and $a$ is proportional to the altitude $h$ from the origin to the plane. In fact, $a = \sqrt{3}h$. Therefore, $x+y+z$ in the integrands can be conveniently replaced by $\sqrt{3}h$, i.e.

$$I_2=\frac{4}{3}\int_{\Omega} \sqrt{3}h dV$$ $$I_3=\frac{4}{3}\int_{\partial\Omega} \sqrt{3}h d\Sigma$$

Now rotate the $z$-axis to be align with the direction (1,1,1) and reparametrize the two balls as,

$$x^2+y^2+z^2=1$$ $$x^2+y^2+(z-\sqrt{3})^2=1$$

All three integrals can then be performed readily along the $z$-axis, albeit in two regions corresponding to the two spherical caps. In the rotated ordinates, the altitude is simply $h=z$, and the three integrals become,

$$I_1 = \int_{\sqrt{3}-1}^{\sqrt{3}/2} \pi[1-(z-\sqrt{3})^2]dz + \int_{\sqrt{3}/2}^1 \pi(1-z^2)dz=\pi\left(\frac{4}{3}-\frac{3\sqrt{3}}{4} \right).$$

$$I_2 = \frac{4}{3}\int_{\sqrt{3}-1}^{\sqrt{3}/2} \pi\sqrt{3}z[1-(z-\sqrt{3})^2]dz + \frac{4}{3}\int_{\sqrt{3}/2}^1 \pi\sqrt{3}z(1-z^2)dz=\pi\left(\frac{8}{3}-\frac{3\sqrt{3}}{2} \right).$$

$$I_3 = \frac{4}{3} \int_{\sqrt{3}-1}^{\sqrt{3}/2} 2\pi\sqrt{3}z\sqrt{1-(z-\sqrt{3})^2}dz + \frac{4}{3}\int_{\sqrt{3}/2}^1 2\pi\sqrt{3}z\sqrt{1-z^2}dz=\pi\left(\frac{2\pi}{3}-\sqrt{3} \right).$$

Answer 2

You intersect two unit balls whose centers have the distance $\sqrt{3}$. Therefore you may take $(\sqrt{3},0,0)$ as center of the second ball. The rim of the resulting lens is in the plane $x={\sqrt{3}\over2}$. We therefore obtain the following volume of the intersection: $${\rm vol}(\Omega)=2\cdot\pi\int_{\sqrt{3}/2}^1(1-x^2)\>dx={16-9\sqrt{3}\over12}\>\pi\ .$$