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I'm trying to find the volume of $f(x,y,z)=y$ inside the cylinder $x^2+y^2=4$ and outside the hyperboloid $x^2+y^2-z^2=1$, and I got the integral

$\int_0^{2\pi} \int_0^y \int_1^\sqrt{1+z^2} rdrdzd\theta$

thank you!

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    $\begingroup$ What did you try so far? $\endgroup$ – Ben373 Mar 4 '18 at 22:04
  • $\begingroup$ I tried to change it in cylindrical coordinates, and I got stuck $r^2=4, \Rightarrow 4-z^2=1, z=\sqrt 3$ $\int_0^{2\pi} \int_\sqrt 3^y \int_2^\sqrt{2-r^2}rdrdzd\theta$ $\endgroup$ – pin_r Mar 4 '18 at 22:07
  • $\begingroup$ There should be no $y$ anywhere in the limits. $y$ is not one of the cylindrical coordinates. $\endgroup$ – Andrei Mar 4 '18 at 22:29
  • $\begingroup$ my mistake, then $\int_0^{2\pi} \int_?^? \int_2^\sqrt{2-r^2}$ $\endgroup$ – pin_r Mar 4 '18 at 23:04
  • $\begingroup$ Integral of the function. And you are forgetting the integrand... $\endgroup$ – Martín-Blas Pérez Pinilla Mar 7 '18 at 20:58
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Cylinder: $$ x^2+y^2 = 4\implies r = 2. $$ Hyperboloid: $$ x^2 + y^2 - z^2 = 1\implies z^2 = r^2 - 1\implies r = \sqrt{z^2 + 1}. $$ Now, you need to find the intersection between cylinder and hyperboloid... Can you continue?

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