Volumetric Calculation of Volcano A

[UrbanXrisis]

A volcano fills the volume between the graphs z=0 and $$\frac{1}{(x^2+y^2)^3}$$, and outside the cylinder $$x^2+y^2=1$$

so I found the z height to be from 0 to 1, the radius from 1 to infinity, and theta to be from 0 to 2pi

$$\int_0 ^{2 \pi} \int_1 ^{\inf}\int _0 ^ {1} r dzdrd \theta$$

I know that this is not correct but I don't know how to set this integral up. any ideas?

 

[HallsofIvy]

That is, your "volcano" is the region between the vertical line r= 1 and the curve z= 1/(x²+ y²⁾3[/sup]= 1/r⁶ rotated around the z-axis. Yes, $\theta$ goes from 0 to $2\pi$ and z goes from 0 to 1 but, for each z, r does not go from 1 to infinity, it goes from the boundary r= 1 to z= r⁶ or r= z^-1/6. Your integral is
$$\int_{\theta= 0}^{2\pi}\int_{z= 0}^1\int_{r=1}^{z^{-1/6}}rdrdzd\theta$$