# Simple Calc 3 Problem

Find the volume of the set $$\left\{ \left(x,y,z\right)\mid-1\le z\le1,4\left(x-\sin z\right)^{2}+\left(y-\cos z\right)^{2}\le1\right\}.$$

I am stuck on this simple problem that is just a matter of setting up the associated triple integral. Any help would be appreciated.

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Hint: for a fixed value $z_0$ of the variable $z$, what is the cross-sectional area of the intersection of your set with the plane $z=z_0$? –  Aaron Sep 13 '12 at 22:17

First choose your order of integration. It looks natural to do the $z$ integration last, as we can see the allowable range is $-1$ to $1$. Then the next one in may as well be $x$ (there doesn't seem any reason to choose one or the other). For this one, $z$ is a fixed value and we need to find the range of $x$. The $(y- cos z)^2$ term can get to but never can be less than $0$, so the range of $x$ has to give $4(x-\sin z)^2 \le 1$ or $x$ ranges from $\sin z - \frac 12$ to $\sin z + \frac 12$. For the inner integral, you consider $x$ and $z$ fixed and find the allowable range in $y$. We have $(y- \cos z)^2\le 1-4(x- \sin z)^2$, so $y$ ranges from $\cos z - \sqrt{1-4(x- \sin z)^2}$ to $\cos z + \sqrt{1-4(x- \sin z)^2}$. The integrand is $1$ so the result is $$\int_{-1}^1dz\int_{\sin z - \frac 12}^{\sin z + \frac 12}dx\int_{\cos z - \sqrt{1-4(x- \sin z)^2}}^{\cos z + \sqrt{1-4(x- \sin z)^2}}1dy$$

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We can actually evaluate this integral without doing any explicit integration. Given any region in $S\subset\mathbb R^3$, the volume is the integral
$$V=\int_S 1 dx dy dz = \int_{\mathbb R} A_z dz$$
where $A_z$ is the cross sectional area of $S$ as we vary $z$.
In this particular case, all the non-empty cross sections are ellipses of height $1$ and width $1/2$ (although the center changes), therefore every cross section has area $\pi/2$. Therefore the volume is
$$\int_{-1}^1 \pi/2 dz = \pi.$$