I am trying to find the volume inside the sphere $x^2 + y^2 + z^2 = 9$, but outside the hyperboloid $x^2 + y^2 – z^2 = 1$. by using a triple integral.
for some reason i just cant seem to come up the bounds of integration for this problem.
To be more precise, its the region lying to the side of the hyperboloid, that wraps around it, creating a sort of donut shape.
[Math] volume inside sphere but outside hyperboloid
integrationmultivariable-calculusvolume
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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!
turns out that I was viewing the figures in the wrong way. As Andrew D. Hwang pointed out the cylinder has infinite height. Since the cylinder ends are open, it engulfs the top and bottom end of the sphere so the endcaps are not included. This explains why the integral:
$$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$
is correct for this problem.
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