Calculate the volume bounded by 2 balls.

Question

I need to calculate the volume bounded by: $$x^2+y^2+z^2\:=\:1,\:x^2+\left(y-1\right)^2+z^2=1$$

My solution: Because the volume I want to calculate is symetric, Ill calculate only one half of it and then double it.

$$\frac{1}{2}\le y\le 1$$ $$x^2+z^2\le 1-y^2$$ Then I do the known substitution: $$x\:=\:r\cos \left(\theta \right)\cos \left(\alpha \right),\:y\:=\:r\sin \left(\theta \:\right)\cos \left(\alpha \:\right),\:z\:=\:r\sin \left(\alpha \right)$$

where the new integral is: $$\int \int \int r^2\sin \left(\alpha \right)$$

when: $$-\frac{\pi }{2}\le \theta \le \frac{\pi }{2},\:0\le \alpha \le \pi ,\:1\le r\le 2$$

A little explanation about how i get $r$:$$\frac{1}{2}\le r\sin \left(\theta \right)\cos \left(\alpha \right)\le \frac{r}{2}\le 1\:$$ because: $$\sin \left(\theta \right)\cos \left(\alpha \right)$$ has maximum when: $$\theta =\frac{\pi }{4},\:\alpha =\frac{\pi }{4}$$

Answer 1

Method1 So you have two spheres, both of which are positioned at $(0,0,0)$ and $(0,1,0)$ with radii $r=1$.

So we can use the formulae $x^2+y^2+z^2=1$ and $x^2+(y-1)^2+z^2=1$ to find their intersection (which you've done already:)

Solving simultaneously gives $$y^2=(y-1)^2 \Longrightarrow y=\frac{1}{2}$$

Now we can use triple integrals to find the volume, but I'm gonna just use the Formula $V=\frac{4}{3} \pi r^3$ : for each sphere this makes their volume $\frac{4\pi}{3}$.

But obviously we have to deal with the overlapping region. We employ the trick that the overlapping spheres are axisymmetric. This lets us calculate the overlapping region by integrating Solids of revolution about the y axis; then it will just be a matter of subtracting this overlap region twice to calculate the real volume contained by the overlapping spheres.

What's the curve that we need to revolve? Let's look at the sphere centred at the origin.

The blue curve is just the projection of the sphere into the xy or yz planes. Let's assume its the xy plane. This curve is described by the equation $$x^2+y^2=1$$

So clearly the solid of revolution we need to consider is of the form

$$\int \pi [x(y)]^2 dy$$.

What is our $x(y)$? Well we can rearrange the equation $$x^2+y^2=1 \Longrightarrow x(y)=\sqrt{1-y^2}$$ to get $x$ in terms of $y$.

What are the limits of integration? This is clearly when the intersection starts and the intersection ends with respect to $y$: Well the overlap for the blue region happens between $\frac{1}{2} \leq y \leq 1$. So we have to evaluate

$$\int_{\frac{1}{2}}^{1} \pi [x(y)]^2 dy=\int_{\frac{1}{2}}^{1} \pi (1-y^2)dy$$.

(if my mental arithmetic serves me well!) this integral gives us a volume of $\frac{5 \pi}{24}$.

By symmetry this is half of the duplicated region, so this means the lens shaped region has volume $\frac{5 \pi}{12}$.

Then it's an adding game: The Total region is $$\frac{4\pi}{3}+\frac{4\pi}{3}-\frac{5 \pi}{12}=\frac{27\pi}{12}=\frac{9\pi}{4}$$

For Interest, I plugged this into mathematica to check:

Answer 2

A simple way through Cavalieri's principle: let we compute the area of the $y$-section, when $y\in\left[0,\frac{1}{2}\right]$. It is the area of a circle, given by $\pi\left(1-(1-y)^2\right) = \pi y(2-y)$. That gives:

$$ V = 2\pi\int_{0}^{\frac{1}{2}} y(2-y)\,dy = \color{red}{\frac{5\pi}{12}}. $$

Answer 3

The volume is made of two spherical caps. To compute the volume of one spherical cap, notice that it is made of infinitesimally short cylinders of radius $r=\sqrt{1-y^2}$ and height $\mathrm dy$. The volume of one cap is obtained by the integral $$\int_{1/2}^1\pi r^2\mathrm dy=\pi\int_{1/2}^1(1-y^2)\mathrm dy=\frac{5\pi}{24}.$$ The volume of the intersection is therefore $5\pi/12$.

Answer 4

So we want to integrate our portion of the sphere as discussed above. We just need to write down the correct inequalities.

So the first inequality we can write down comes from what we know about $y$ i.e.

$$ \frac{1}{2}\leq y \leq 1 $$

The volume bounded by the sphere is also defined by $$x^2+y^2+z^2 \leq 1 $$

Rearranging for $z$ we attain

$$-\sqrt{1-x^2-y^2} \leq z \leq \sqrt{1-x^2-y^2} $$

So we have our limits of integration for $y$ and $z$.

Now to find the limits of $x$, we need to find a graph that depends on $x,y$ but independent of $z$; take the level set when $z=0$ i.e.

$$x^2+y^2 \leq 1 $$ Then we can find the inequalities for $x$ by rearranging i.e.

$$-\sqrt{1-y^2}\leq x \leq \sqrt{1-y^2} $$

Then we integrate as usual (for clarity i'll integrate in order of $z,x,y$ (which can be done by rules of multivariable calculus):

$$Vol(V)=\iiint_{V} dV= \int_{y=\frac{1}{2}}^{y=1} \int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}} \int_{z=-\sqrt{1-x^2-y^2}}^{z=\sqrt{1-x^2-y^2}} dzdxdy $$

$$=\int_{y=\frac{1}{2}}^{y=1} \int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}} 2\sqrt{1-x^2-y^2} dxdy $$

$$=\int_{y=\frac{1}{2}}^{y=1} 2 \cdot \frac{1}{2} \pi (1-y^2) dy $$

$$=\frac{5\pi}{24} $$

Would one go ahead and try to convert into spherical polars? probably not; the region isn't easily described at $y=\frac{1}{2}$ i.e. the infinitesimal section in spherical coordinates is $r^2 sin(\phi) dr d\theta d\phi$ (so I won't bother). But cylindrical coordinates is perfect for axisymmetric volumes so we could use it here.

To make things easier, I'll relabel $x \rightarrow x$, $y \rightarrow z$ and $z \rightarrow y$ so the $z$-axis skewers the cap we are working on.

Then in polar coordinates, $x=r \cos(\theta), y=r \sin(\theta),z=z$

and we can write the region by $$z^2+r^2\leq 1, \frac{1}{2} \leq z \Longrightarrow \frac{1}{2} \leq z\leq \sqrt{1-r^2} $$ $$0 \leq r \leq 1 $$ and $$0 \leq \theta \leq 2\pi $$ Then $$Vol(V)=\iiint_{V} dV = \int_{\theta=0}^{2\pi} \int_{r=0}^{1} \int_{z=\frac{1}{2}}^{\sqrt{1-r^2}} rdzdrd\theta $$ Which should give the correct answer