# Volume of the solid intersecting 3 spheres

Let the next three spheres: \begin{array}{lcccl} S_1 : &(x-1)^2 &+ &y^2 &+ &z^2 &=1, \\ S_2 : &x^2 &+ &y^2 &+ &z^2 &=1, \\ S_3 : &(x+1)^2 &+ &y^2 &+ &z^2 &=1. \end{array}

I have to calculate the volume of the solid inside $S_2$ and outside $S_1$ and $S_3$.

Can you help me to determine the bounds of each integral if I have to use the cylindrical coordinates?

• It might be easier to compute the volumes of the portions of $S_{1}$ and $S_{3}$ that are inside $S_{2}$ and then subtract that total from the volume of $S_{2}$ (which is, of course, $\frac{4}{3}\pi$). Nov 23, 2015 at 19:40

You tag this as multivariable-calculus, implying you want a double or triple integral, but this can be done with a single integral.

You included the following diagram in the original version of your question. This is the intersection of your solid with the $xy$-plain: i.e. the graphs of your inequalities given $z=0$. (In the inequalities, the second equation has less-than-or-equal-to, while the others have greater-than-or-equal-to.) Your desired solid is the rotation of the upper half of this diagram about the $x$-axis. And that is double the rotation of the upper-right quarter of the diagram about the $x$-axis. The lower bound of $y$ of that upper-right quarter in the first quadrant is

$$y_1=\sqrt{1-(x-1)^2}=\sqrt{2x-x^2}$$

and the upper bound is

$$y_2=\sqrt{1-x^2}$$

And you easily see that the bounds of $x$ are $0\le x\le \frac 12$.

So use the washer method to calculate that rotation of that upper-right quarter about the $x$-axis, then double that.

The answer is \begin{align} V&=2\cdot\pi\int_0^{1/2}(y_2^2-y_1^2)\,dx \\[2 ex] &=2\pi\int_0^{1/2}\left[\left(\sqrt{1-x^2}\right)^2 -\left(\sqrt{2x-x^2}\right)^2\right]\,dx \\[2 ex] &=2\pi\int_0^{1/2}[1-2x]\,dx \\[2 ex] &=2\pi[x-x^2]_0^{1/2} \\[2 ex] &=2\pi\left[\left(\frac 12-\frac 14\right)-(0-0)\right] \\[2 ex] &=\frac{\pi}2 \end{align}

This is how to solve your problem with cylindrical coordinates.

As in the previous section, we want to double the volume of the right half of the solid ($x\ge 0$) which itself comes from a full rotation of the region in the 1st quadrant. There we have $0\le x\le \frac 12$. The values of $r$ are the values of $y$ in the 1st quadrant of the diagram: $\sqrt{2x-x^2}\le r\le \sqrt{1-x^2}$. We want a full rotation of the 1st quadrant, so $0\le\theta\le 2\pi$.

So we want double the volume of the right half, integrating over cylindrical coordinates.

\begin{align} V &= 2\iiint r\,dr\,d\theta\,dx \\[2 ex] &= 2\int_0^{1/2}\int_0^{2\pi}\int_{\sqrt{2x-x^2}}^{\sqrt{1-x^2}}r\,dr\,d\theta\,dx \\[2 ex] &= 2\int_0^{1/2}\int_0^{2\pi}\left[\frac 12r^2\right]_{\sqrt{2x-x^2}}^{\sqrt{1-x^2}}\,d\theta\,dx \\[2 ex] &= 2\int_0^{1/2}\int_0^{2\pi}\frac 12\left[\left(\sqrt{1-x^2}\right)^2-\left(\sqrt{2x-x^2}\right)^2 \right]\,d\theta\,dx \\[2 ex] &= \int_0^{1/2}\int_0^{2\pi}(1-2x)\,d\theta\,dx \\[2 ex] &= \int_0^{1/2}2\pi(1-2x)\,dx \\[2 ex] &=2\pi[x-x^2]_0^{1/2} \\[2 ex] &=2\pi\left[\left(\frac 12-\frac 14\right)-(0-0)\right] \\[2 ex] &=\frac{\pi}2 \end{align}

You see that the answer is the same as that of the washer method. Indeed, the last few lines are also the same, and much of the working is very similar.

Here is a derivation of the bounds on the integrals.

If we use cylindrical coordinates, with the axis of the cylinder along the $x$-axis, we will get $y^2+z^2=r^2$. Now we can rewrite your "equations" and the "inside/outside" requirements as three inequalities using $x$ and $r$:

$$(x-1)^2+r^2\ge 1, \quad x^2+r^2\le 1, \quad (x+1)^2+r^2\ge 1$$

We can see that those are symmetric in $x$, so let's just look at $x\ge 0$. We can rewrite the first two inequalities to get bounds on $r$, remembering that $r$ is positive. The first gives

$$r\ge \sqrt{2x-x^2}$$

and the second gives

$$r\le \sqrt{1-x^2}$$

Reversing the first inequality and adding the first two together gives

$$x^2+r^2+1\le (x-1)^2+r^2+1$$

$$x\le\frac 12$$

We already said $x\ge 0$, so the bounds on $x$ are $0\le x\le \frac 12$.

There is no restriction at all on $\theta$, but we don't want to repeat any points, so we use one full circle $0\le\theta\le 2\pi$.

And those are the bounds on the integrals. Note that using only $x\ge 0$ meant we could ignore the third inequality.

• I didn't learn the washer method yet. So, I suppose this is why I have to calculate it with multiple integrals. And, maybe, after that, I will have to calculate it with a density function. Nov 23, 2015 at 19:46
• The washer method is taught in 1st year calculus and is quite easy: follow the link I gave. Unless you are explicitly told not to use this method, I recommend that you use it here: it is easier than a multiple integral, and your teacher would expect that you have already seen it. Nov 23, 2015 at 19:56
• Thanks for this answer. Now, I can say I know what's the washer method. But, if I want to calculate it with a density function, I will have to consider a multiple integrals... How can I determine the bounds? Nov 23, 2015 at 19:57
• If the density function method is required, you should state that in your main question. The bounds on $x$ are clearly $0\le x\le \frac 12$, but the "slice" of $yz$ for a given $x$ is a ring or "washer". Putting that into a double integral would require multiple pieces: above and below the inner hole, and to the left and right of it. Is that really what you want? Cylindrical coordinates would work well, with $r$ having the bounds $y_1$ and $y_2$ as in my answer, and $0\le\theta<2\pi$. Can you use cylindrical coordinates? Nov 23, 2015 at 20:08
• Yes...Yes... I learn that recently. How can I convert this problem into cylindrical coordinates? Nov 23, 2015 at 20:11

As an alternative approach, if you happen to know some handy formulas:

The volume in $S_2$ that is outside $S_1$ and $S_3$ is equal to the volume of $S_2$ (which is $\frac{4\pi}{3}$) less the volume of four endcaps, each of "height" $1/2$. The expression for a volume of a spherical cap of height $h$ in a unit sphere is

$$V_\text{cap} = \frac{\pi h^2}{3}(3-h)$$

With $h = 1/2$, we have $V_\text{cap} = \frac{5\pi}{24}$, so the desired volume is

$$V = \frac{4\pi}{3}-4 \times \frac{5\pi}{24} = \frac{\pi}{2}$$