volume between sphere $(x+1)^2+y^2+z^2=4$ and cylinder $x^2+y^2=1$.

Question

I have the sphere given by $(x+1)^2+y^2+z^2=4$ and the cylinder $x^2+y^2=1$, and I am looking for the volume enclosed by the intersection of the two surfraces. Using cylindrical coordinates, $\theta$ runs from $0$ to $2\pi$. I want to construct the boundaries for $z$ and $r$. I think that $r$ runs from $0$ to $1$. Then $z$ depends on $R$, but how do I construct this $z(r,\theta)$, I get $z=\sqrt{4-r^2+2r\cos\theta}$. Is this correct? Then, how do I evaluate the integral $$ \int_{0}^1\int_{0}^{2\pi}\sqrt{4-r^2+2r\cos\theta}d\theta dr? $$ Would it be better to go for spherical coordinates?

Answer 1

It is important to check that the disk $x^2+y^2\le 1$ lies entirely inside the disk $(x+1)^2+y^2\le 4$, so you are right that $\theta$ goes from $0$ to $2\pi$ and $r$ goes from $0$ to $1$. However, you screwed up your algebra solving for $z$ on the upper hemisphere. Shouldn't it be $z=\sqrt{3-r^2-2r\cos\theta}$? And, of course, you need to double your integral to get the volume below the $xy$-plane, as well. Last, you're missing the $r$ in $r\,dr\,d\theta$.

The integral is going to be very difficult to do. What if you change your viewpoint and consider the sphere $x^2+y^2+z^2=4$ and the cylinder $(x-1)^2+y^2=1$ instead? Then you get an easy integral to do, provided you do the usual order $dr\,d\theta$.

Answer 2

Re-center the sphere at origin and use the cylindrical coordinates to write the two surfaces as,

$$r^2+z^2=4,\>\>\>\>\>r=2\cos\theta$$

The volume integral can then be set up as,

$$V=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}dz\>rdr\>d\theta$$ $$=2\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}{\sqrt{4-r^2}}rdr\>d\theta =\frac{16}{3}\int_{-\pi/2}^{\pi/2}(1-|\sin^3\theta|)d\theta=\frac{16}3\left(\pi-\frac43\right)$$