Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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 $$\int_{-1}^1dz\int_{\sin z - \frac 12}^{\sin z + \frac 12}dx\int_{\cos z - \sqrt{1-4(x- \sin z)^2}}^{\cos z + \sqrt{1-4(x- \sin z)^2}}1dy$$

share|improve this answer

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

$$ \int_{-1}^1 \pi/2 dz = \pi.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.