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$K=\{ (x,y,z)\in\mathbb R^3 | z\in[-\frac{\pi}{2}, \frac{\pi}{2}], x^2+y^2 \le \cos^2(z) \}$

Compute the 3 dimensional volume of K.

So $\int_{\mathbb R^3} \mathbb1_K(x,y,z) d\lambda(x,y,z)$ but at this point I am stuck, I know there are polar and spherical coordinates and change of variables but I can't see the proper subsitution. One could try something like

$(x,y,z)=\Phi(z,\theta)=(\cos(z)\cos(\theta), \cos(z)\sin(\theta), z)$ but I don't think that this is the desired solution.

Any suggestions?

1 Answers 1

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It's one loop in the surface of revolution of cosine function,

\begin{align*} V &= \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{\cos z} r\, dr \, \, d\theta \, dz \\ &= 2\pi \int_{-\pi/2}^{\pi/2} \int_{0}^{\cos z} r\, dr \, dz \\ &= 2\pi \int_{-\pi/2}^{\pi/2} \left[ \frac{r^2}{2} \right]_{0}^{\cos z} dz \\ &= \pi \int_{-\pi/2}^{\pi/2} \cos^2 z \, dz \\ &= \pi \int_{-\pi/2}^{\pi/2} \frac{1+\cos 2z}{2} \, dz \\ &= \pi \left[ \frac{z}{2}+\frac{\sin 2z}{4} \right]_{-\pi/2}^{\pi/2} \\ &= \frac{\pi^2}{2} \end{align*}

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    Thank you, I should have seen this. But are you sure that $r$ goes from $0$ to $\cos^2(z)$? Shouldn't it be $\cos(z)$? – 2017-02-16
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    Ah! Sorry for the mistake, correct very soon! – 2017-02-16
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    No hurry, it should be even nicer that way :) – 2017-02-16
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    $\frac{\pi^2}{2}$ ? – 2017-02-16
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    Yes! Hope there's no bug anymore. – 2017-02-16