Question: I am not getting the correct answer. How do I get the solution (and why does my solution not work?)

Find volume that lies inside both spheres:

\begin{align} A: 4 &= (x+2)^2 + (y-1)^2 + (z+2)^2\\ B: 4 &= x^2 + y^2 + z^2\\ \end{align}

My solution gives the answer $V \approx 11$, whereas Chegg.com gives the answer $\frac{11}{24}\tau$, where $\tau = 2\pi$. However, I don't like their method because they pull the equation for the volume of the cap of a sphere out of thin-air with no explanation. I was hoping to find a more intuitive answer.

  1. The distance between the spheres is $d = \sqrt{2^2 + 1^2 + 2^2} = 3$, so I can replace these spheres with two similar spheres 3 units apart, lying along the $x$-axis.
  2. I project the spheres into the $xy$-plane. The line of intersection is $x = 1.5$.
  3. Next, I integrate over $x: [1.5 , 2]$ by volume of rotation using the disc method: $ V = \tau \int f(x) \ dx$.

    This should give the volume of half of the volume inside both spheres. If I double the integral, I should get the entire volume, so:

$$V = 2\tau\int_{1.5}^{2} \sqrt{x^2 - 4}\ dx$$

Let: $x=2\sin\theta$. Then: $dx = 2\cos\theta\ d\theta$.

$$V =2\tau\int_{\arcsin(3/4)}^{\tau / 4} \sqrt{(2\sin\theta)^2 -4} \cdot2\cos\theta\ d\theta$$

Integrate by factoring out 4 from squareroot, then use the Double-Angle Formula.

$$V=4\tau(\theta + \tfrac{1}{2}\sin2\theta)\Big]_{\arcsin(3/4)}^{\tau/4}\approx 11 \neq \frac{11}{24}\tau$$

What did I do wrong?


Your instincts are right - your math is wrong. The volume is proportional to the integral of the square of the function. By the disk method, I get that the volume is

$$2 \pi \int_{3/2}^2 dx \, y^2=2 \pi \int_{3/2}^2 dx \, (4-x^2)$$

This evaluates to $2 \pi (11/24)$.

  • $\begingroup$ Wow, thanks. I can't believe I forgot. $\endgroup$ – JDG Aug 19 '13 at 16:13
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    $\begingroup$ +1 Excellent Ron. I am thinking the way we could find the volume by using triple integrals. I don't think we can find this way. Your approach is very effective for this kind of volume. Thanks for sharing me that. :-) $\endgroup$ – Mikasa Aug 20 '13 at 5:39
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    $\begingroup$ This can be solved without explicit integration. See, for example, mathworld.wolfram.com/Sphere-SphereIntersection.html $\endgroup$ – Carl Witthoft Feb 1 '16 at 14:27

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