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$0\lt a \lt b$ and $c\gt 0$ and I want to calculate $\lambda ^3(S)$.

I think I have to use some transformation but I already tried transforming into polar coordinates and it didn't work. Any tipps on what transformation to use? Thanks in advance!

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I'll assume that $a$, $b$ and $c$ are positive. The set $S$ has a cylindrical symmetry around the $z$ axis, so it can be convenient to change the coordinates as \begin{equation} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} \rho\cos\theta\\ \rho\sin\theta\\ z \end{pmatrix}, \end{equation} with $(\rho,\theta,z)\in(0,+\infty)\times(0,2\pi)\times\mathbb{R}$. Let $g$ be the transformation from cylindrical to rectangular coordinates: it is easy to see that \begin{equation} g^{-1}(S)=\{z>0,a^2<\rho^2+z^2

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The volume is then \begin{equation} \lambda^3(S)= \int_0^{2\pi}\mathrm{d}\theta\int_U\rho\,\mathrm{d}\lambda' \end{equation} where $\lambda'$ is the Lebesgue measure on the plane and \begin{equation} U=\{(\rho,z)\in\mathbb{R}^2\colon a^2<\rho^2+z^2< b^2, 00$) \begin{equation} \eta<\arctan c, \end{equation} so \begin{equation} h^{-1}(U)=\{(r,\eta)\in(0,+\infty)\times(0,2\pi)\colon a

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    Ah ok so i have to switch coordinates two times. Thanks a lot i didn't think of that.2017-01-29
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    Actually I made a mistake, I switched coordinates too soon the second time. I corrected my answer.2017-01-29