I have done an exercise in two different ways but I have obtained two different results and I can't understand what's wrong. Please, help me.
Given: $V=\{(x,y,z)\in R^3: x^2+y^2+z^2\leq 1, \frac{1}{3}\leq z \leq \frac{1}2\}$
Calculate $\iiint_V z \sqrt {x^2+y^2} dxdydz$
My FIRST attempt at solution: In spherical coordinates we have:
$x= r \cos \psi \cos \phi$
$y=r \cos \psi \sin \phi$
$z=r \sin \psi$
so, $\sqrt {x^2+y^2}$ becomes $r \, |\cos \psi |$
from the condition on $z$ I obtain $\displaystyle \frac{1}{3 \sin \psi}<r<\frac{1}{2\sin \psi}$
and the integral I have to calculate begins:
$\iiint r \sin \psi \cdot r |\cos \psi| \cdot r^2 \cos \psi \, d\phi d\psi dr= \\ \iiint r^4 \cos^2 \psi \sin \psi \, d\phi d\psi dr= \\ 2 \pi \int (\int_a ^b r^4dr) \cos^2 \psi \sin \psi d\psi$
and I have obtained
$=\displaystyle \frac{2}{5}\pi \int_{\arcsin 1/3}^{\pi/2} \left(\frac1{2^5 sin^5 \psi}-\frac{1}{3^5 \sin ^5 \psi}\right)\cos^2 \psi \sin \psi \, d\psi$
WolframAlpha says:
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My SECOND attempt at solution was in cylindrical coords:
$\int_{1/3}^{1/2} \int_{0}^{2\pi} \int_{0}^{\sqrt{1-z^2}} z \rho \cdot \rho \, d\rho d\theta dz= $
Wolfam says:
I probably made a mistake somewhere and I can't find it! Please, check it out!
This is a horrible mess in spherical coordinates; the cylindrical coordinates are clearly more suitable. But if you insist on doing it, here's your mistake: The upper bound for $r$ is $1/(2\sin\psi)$ only in part of the range; for $\psi\lt\arcsin\frac12$ the upper bound for $r$ is $1$; so you have to split the integral over $\psi$ into two parts. Think of rays piercing the slab of the sphere from the origin; some emerge from the slab on the surface of the sphere, and some on the upper flat surface of the slab.