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The question says to evaluate the given triple integral in cylindrical coordinates over the region $W$. Here's the integrand and the region: enter image description here

I'm confused about this question because I keep getting $0$. Here's the triple integral in polar coordinates that represents the region given in question $9$:

Note that I use $x=r\cos\theta$, $y=r\sin\theta$, and $z=z$ to simplify the integrand $f(x,y,z)=x^2+y^2+z^2$.

$$\int_{-1}^1\int_{\pi/4}^{3\pi/4}\int_{0}^4 r(r^2+z^2) \ dr\ d\theta \ dz$$

After integrating the inner intergral we end up with positive powers of $r$. As a result, the function evaluated at $1$ minus the function evaluated at $-1$ ends up being $0$, and the integration stops there. The order of integration turns out not to matter: I always get this result.

Yet the text shows a non-zero number as the answer. What am I missing?

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The answer is non-zero. I suspect you're calculating the integrals incorrectly, since your setup looks great. We have \begin{align*} \int_{-1}^1\int_{\pi/4}^{3\pi/4}\int_{0}^4 r\big(r^2+z^2\big) \ dr\ d\theta \ dz &=\frac{\pi}{2}\int_{-1}^1\int_0^4\big(r^3+rz^2\big)\,dr\,dz\\ &=\frac{\pi}{2}\int_{-1}^1\left(\frac{r^4}{4}+\frac{r^2z^2}{2}\right)\bigg|_{0\to 0}^4\,dz\\ &=\frac{\pi}{2}\int_{-1}^1\left(64+8z^2\right)dz\\ &=\pi\int_{0}^1\left(64+8z^2\right)dz\quad\text{(even function on symmetric interval)}\\ &=\pi\left(64z+\frac{8z^3}{3}\right)\bigg|_{0}^{1}\\ &=\frac{200\pi}{3}. \end{align*}

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  • $\begingroup$ Thank you very much, this helps a lot! I was going wrong in multiple places during my calculation. I appreciate the help! $\endgroup$ – James Ronald Oct 7 at 21:33
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Not true... $$ \int_{0}^4 r(r^2+z^2)dr=\left[\frac{r^4}{4} + \frac{r^2 z^2}{2}\right]_{0}^4 = \frac{4^4}{4} + \frac{4^2 z^2}{2}=64+8z^2 $$

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  • $\begingroup$ Sorry, I meant for that $dz$ to be a $dr$; I've fixed it now. So you'd actually be evaluating $r^4 + (r^2z^2)/2$ over $1$ and $-1$ with respect to $r$. The positive powers of $r$ in that integrand make it so that the entire thing has the same value whether you're evaluating for $1$ or $-1$ $\endgroup$ – James Ronald Oct 7 at 14:07
  • $\begingroup$ @JamesRonald When you exchange the order of integration you must be consistent in the integration limits. If you start with $dr$, the integration limits are $0,4$, not $-1,1$. $\endgroup$ – PierreCarre Oct 7 at 14:09
  • $\begingroup$ Indeed, I fixed that as well, thank you. Do you know how to solve this now that we're evaluating with respect to $r$ first? $\endgroup$ – James Ronald Oct 7 at 14:10
  • $\begingroup$ @JamesRoland The result is $\frac{200 \pi}{3}$, regardless of the integration order. $\endgroup$ – PierreCarre Oct 7 at 14:11
  • $\begingroup$ Right, but how do we get that...? $\endgroup$ – James Ronald Oct 7 at 14:13

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