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I'm trying to find the volume bounded by the graphs of $z = 0$ and $z = h$, outside of the cylinder $x^2 + y^2 = 1$, and inside the hyperboloid $x^2+y^2-z^2 = 1.$ I have tried to use cylindrical coordinates, and I got the integral $$\int_0^{2\pi}\int_1^{\sqrt{1+h^2}}\int_0^hr dz dr d\theta$$ However the answer is $h^3\pi/3$, and the integral above computes to $h^3\pi$. I have graphed the region several times and I don't know what I'm doing wrong. Any help will be greatly appreciated.

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This answer comes a little later than you'd wanted it - sorry! There's a small mistake with your limits. Even though you can turn x^+y^2=1-z^2 into r=sqrt(1+z^2), you can't swap in h for z just because your top boundary is z=h. Z (a variable) isn't always h (a constant in this case); just at that top edge. So keep it as z. I set my bounds like this: 0 to 2pi, 0 to h, 1 to sqrt(1+z^2). Integrate that in terms of r-dr-dz-dtheta and you should get your answer. Best of luck!

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  • $\begingroup$ Thanks. This really helped! $\endgroup$ – Vishwa Iyer May 5 '14 at 0:27

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