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I have a tricky triple integral and I'm not sure if what I'm doing is correct. The problem is as follows:

"Let $K \subset \mathbb{R^3}$ be the body consisting of the points $(x,y,z)$ that satisfy the conditions

\begin{equation} z(1-z) \leq \sqrt{x^2+y^2} \leq 2z(1-z),\ x \geq y \end{equation}

Calculate the volume of $K$."

I decided to use cylindrical coordinates. $z$ remains the same, and from the equation we see that $z$ goes from zero to 1. For $r$, we get the inequality $z(1-z) \leq r \leq 2z(1-z)$. For $\theta$, we have $r\ cos\ \theta \geq r\ sin\ \theta$, so $tan\ \theta \leq 1$, so $ 0\leq \theta \leq \frac{\pi}{4}$.

So we would then get $\iiint_K\ 1 = \int_0^{\frac{\pi}{4}}\int_0^1\int_{z(1-z)}^{2z(1-z)} r\ dr\ dz\ d\theta$. Would this be correct is my question? Thanks in advance!

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  • $\begingroup$ I think you mean $\tan \theta \le 1$. $\endgroup$ – mattos Jun 26 '17 at 9:58
  • $\begingroup$ You are right. I edited my question. $\endgroup$ – user444389 Jun 26 '17 at 10:00
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everything is correct except $x\ge y$ is not only $0\le\theta\le\frac\pi4$, you also have $\frac54\pi\le\theta\le2\pi$. When you divide by $\cos\theta$ you get $\tan\theta\le 1$ only when $\cos\theta>0$.

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  • $\begingroup$ Ah okay. So the correct integral would be $\iiint_K\ 1 = \int_0^{\frac{\pi}{4}}\int_0^1\int_{z(1-z)}^{2z(1-z)} r\ dr\ dz\ d\theta + \int_{\frac{5 \pi}{4}}^{2\pi}\int_0^1\int_{z(1-z)}^{2z(1-z)} r\ dr\ dz\ d\theta$. $\endgroup$ – user444389 Jun 26 '17 at 10:18
  • $\begingroup$ yes this is the correct one $\endgroup$ – Gio67 Jun 26 '17 at 10:20

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