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Problem (more here and the problem XIV.6:6 here on page 976)

Integrate $\int_A z dx dy dz$ in cylinder-coordinates when $$A=\{(x,y,z)\in\mathbb R^3 | x^2+y^2 \leq z \leq \sqrt{2-x^2-y^2}\}.$$

So $r^2=x^2+y^2$ and

$$\int_0^{2\pi} \int_a^b \int_{r^2}^{\sqrt{2-r^2}} rz dz dr d\rho$$ but what $a$ and $b$ are? I know for the max -case $0\leq z \leq \sqrt{2}$ and $0\leq r \leq 2^{0.25}$ if $z,r\in\mathbb R$. $$\int_0^{2\pi} \int_0^{2^{0.25}} \int_{r^2}^{\sqrt{2-r^2}} rz dz dr d\rho,$$ just this easy? With this specification, I got $\pi \left( \sqrt{2} - 0.5-\frac{1}{6} 2^{1.5} \right)$ as a solution. But if wrongly specified, it is not right. Perhaps, I need to find an expression for the $r$ in terms of the angle $\rho$ -- somehow from $x=r \cos(\rho)$, $y=r\sin(\rho)$ and $z=z$? I may be misunderstanding here the exercise, perhaps I am just over-engineering...

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With Joseki's notice, I got $\frac{7}{12} \pi$ as a solution (haven't verified). – hhh Apr 30 '12 at 9:24
up vote 1 down vote accepted

I don't know what you mean by "the max-case". You need to integrate over the entire range of positive $r$ values for which the lower bound for $z$ is lower than the upper bound. The bounds are equal at $r^2=\sqrt{2-r^2}$, and thus $r=1$, so the integral with respect to $r$ should be over $[0,1]$.

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Given $x^2+y^2 \leq z \leq \sqrt{2-x^2-y^2}$. In cyl-coords, $r^2 =x^2+y^2$ so $r^2 \leq z \leq \sqrt{2-r^2}$ so $r^4 \leq 2-r^2$ so $r^4+r^{2} -2\leq 0$ so $r^2=\pm 1$. And because $r\geq 0$ we have $0\leq r \leq 1$, yes you are right. I had a stupid mistake with this $r^2=\sqrt{2-r^2}$ (I for some reason thought that the $r$ was different on both sides but apparently not), now corrected thanks. – hhh Apr 30 '12 at 9:14

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