Volume of a solid in $xyz$-space

Question

Calculate the volume of the solid in $xyz$-space bounded by the surfaces $$ \displaystyle z=\frac{1}{x^2+y^2+1} and\:z=\frac{1}{x^2+y^2+4} $$

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I haven't done triple integrals for a long time. But it looks like we can use a coordinate change (cylindrical or spherical?) to make it simpler. And help would be appreciated!

Answer 1

Let $x=r \sin \theta,y=r \cos \theta$, and the volume is, removing $\theta$ having noticed that the graph is unchanged after a rotation about the $z$ axis,

$$V=\int_{r \ge 0}2 \pi r\left(\frac{1}{r^2+1}-\frac{1}{r^2+4} \right)dr= \int_{u \ge 0} \pi \left(\frac{1}{u+1}-\frac{1}{u+4} \right)dr= \pi \int_1^4\frac{1}{u}du=\pi\ln(4).$$

Note that I've used that the volume of a cylinder with height $z$, thickness $dr$ and radius $r$ is $2 \pi r z dr$.

Answer 2

You can use cylindrical coordinates. Then $x^2+y^2=r^2$. You now have a solid of revolution, revolving a curve that looks like this around the $x$ (or $r$) axis. A plot is . Now you should be able to integrate this. I would suggest shells are easier.