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I am asked to find the volume of the solid inside both $x^2+y^2+z^2=a^2$ and $(x-a/2)^2+y^2=(a/2)^2$ using a triple integral on cylindrical coordinates.

I have this triple integral setup, which makes so much sense to me. Nonetheless, the result of the integral is not $\frac{2a^3}{9} \left( 3 \pi - 4 \right)$, as it should be according to the book's answer (Larson, 10th ed.)

$$\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a \cos(\theta)} \int_{- \sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}} \ r \ dzdrd\theta$$

Am I missing something?

Thank you.

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  • $\begingroup$ Check the integration. I did it with that limits and it results in the given solution. $\endgroup$ Commented May 18, 2017 at 19:11
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    $\begingroup$ After the second integration I have $\int_{-\frac {\pi}{2}}^{\frac{\pi}{2}} \frac 23 a^3(1-|\sin^3\theta|) \ d\theta$ The need for the absolute value on the $\sin$ function may have slipped by $\endgroup$
    – Doug M
    Commented May 18, 2017 at 19:16

1 Answer 1

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In order to better understand the solid region $B$ enclosed by both surfaces in order to better evaluate our limits of integration, we can visualize the surfaces as follows:

enter image description here

In cylindrical coordinates, we can express the cylinder as: $$ r = a \cos \theta \qquad (0 \leq \theta \leq \pi)$$ and the sphere as: $$ r^2 + z^2 = a^2$$

We see that, as we traverse all the $(r,\theta)$ or $(x,y)$ points within the cylinder's projection on the $x$-$y$ plane (a circle), the values of $z$ we want are between $\pm\sqrt{a^2-r^2}$

Finally, we recall that: $$ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz $$

So: \begin{align*} \iiint_B \, dV &= \int_{0}^{\pi} \int_0^{a \cos \theta}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\ &= \int_{0}^{\pi} \int_0^{a \cos \theta} 2\int_{0}^{\sqrt{a^2-r^2}} r \, dz \, dr \, d\theta \\ &= \int_{0}^{\pi} \int_0^{a \cos \theta} 2r\sqrt{a^2-r^2} \, dr \, d\theta \\ &= \int_{0}^{\pi} \left[ -\frac{2}{3}(a^2-r^2)^{3/2} \right]_0^{a \cos \theta} \, d\theta \\ &= \frac{2}{3}\int_{0}^{\pi} (a^3 - a^3 \sin^3 \theta) \, d\theta \\ &= \frac{2a^3}{3}\int_{0}^{\pi} (1 - \sin^3 \theta) \, d\theta \\ &= \frac{2a^3}{12}\int_{0}^{\pi} (4 - 3\sin \theta + \sin 3\theta) \, d\theta \\ &= \frac{a^3}{6} \left[4\theta + 3 \cos \theta - \frac{1}{3}\cos 3\theta \right]_0^{\pi} \\ &= \frac{a^3}{6} \left(4\pi - 3 + \frac{1}{3} - 3 + \frac{1}{3} \right) \\ &= \frac{a^3}{6} \left(4\pi - \frac{16}{3} \right) \\ &= \frac{2a^3}{9} \left(3\pi - 4 \right) \end{align*}

The problem that you probably ran into is parametrizing from $\theta = -\pi/2$ to $\theta = \pi/2$. This is a valid parametrization, but when you reach this point: $$ L = \frac{2}{3}\int_{-\pi/2}^{\pi/2} \left[a^3 - (\sin^2 \theta)^{3/2} \right] \, d\theta $$

We have that: $$ (\sin^2 \theta)^{3/2} \neq \sin^3 \theta $$ for $\sin \theta < 0$, which occurs for $-\pi/2 \leq \theta < 0$.

So we choose a parametrization where we keep $\sin \theta \geq 0$, which is why we use $0 \leq \theta \leq \pi$.

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  • $\begingroup$ Wish I could give you +5 for flagging the likely error as well. I was working on a variation with the cylinder lying on the y-axis and suspected this was the issue (but in my case for $\cos\theta$ on $[0,\pi]$). Thank you for the clear presentation & extra attention to probable issues. $\endgroup$
    – Rax Adaam
    Commented Jun 1, 2019 at 17:38
  • $\begingroup$ No problem @RaxAdaam. Glad to be of help. $\endgroup$ Commented Jul 16, 2019 at 17:49

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