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If S is the solid obtained by intersecting the ball $x^2+y^2+z^2\le4$ and $x^2+z^2\le1$

1) How do I show that S is Jordan measurable? Can I simply say the following: "Clearly S is bounded, and the boundary is a union of smooth curves in $R^3$. Since smooth curves have zero content, the boundary is 0."

2) Also for obtaining the volume of S, I am not sure how to calculate the bounds... Do I need the intersections?

Many thanks!

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I do hope that the OP has figured this out in more than four years, but since I found the question future people will also, so just for future reference here we go.

1) I'd indeed use a theorem like that, because otherwise proving this would get pretty complicated which is most of the time not intended, except for the 'do it once in your lifetime' sort of question.

2) First, think of what this is. It is a cylinder, which is bounded on two sides by the sphere. So it may be a good idea to use cylindrical coordinates $r,\theta,y$ where $r$ denotes the radius of the cylinder, $\theta$ the angle to some reference line perpendicular to the cylinder axis and $y$ the height (because the cylinder axis is the $y$-axis).

We now figure out what the boundaries will be for $r,\theta,y$ with the volume we have.

  • $0 \le \theta \le 2\pi$ because this is a full cylinder.
  • $0 \le r \le 1$ because the radius of our cylinder is one.
  • $-\sqrt{4-r^2} \le y \le \sqrt{4-r^2}$. Now this needs some more explanation.

Our cylinder is bounded on two sides by the boundary of the sphere. The boundary is given by $x^2 + y^2 + z^2 = 4$, so solving for $y$ gives $y = \pm \sqrt{4-x^2-z^2}$. But we know with Pythagoras that $r^2 = x^2 + z^2$ (remember the cylinder is around the $y$-axis), so hence the boundaries.

Now we know the volume will be (you could switch $r$ and $\theta$ if you want) $$ \int_0^{2\pi} \int_0^1 \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \, \mathrm{d} y \mathrm{d} r \mathrm{d} \theta , $$ and we are left with some calculations.

We now have

\begin{align*} \text{Vol}(S) &= \int_0^{2\pi} \int_0^1 \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \, \mathrm{d} y \mathrm{d} r \mathrm{d} \theta \\ &= \int_0^{2\pi} \int_0^1 2r \sqrt{4-r^2} \, \mathrm{d} r \mathrm{d} \theta \end{align*} In order to calculate this, we can substitute $u=4-r^2$. We then get $\mathrm{d} u = -2 \mathrm{d} r$, so $\mathrm{d} r = \frac{\mathrm{d} u}{-2r}$. So now, (don't forget to substitute the integral boundaries!) \begin{align*} \text{Vol}(S) &= \int_0^{2\pi} \int_0^1 2r \sqrt{4-r^2} \, \mathrm{d} r \mathrm{d} \theta \\ &= \int_0^{2\pi} \int_4^3 -\sqrt{u} \, \mathrm{d} u \mathrm{d} \theta \\ &= \int_0^{2\pi} \frac{16}{3}-2 \sqrt{3} \, \mathrm{d} \theta \\ &= \frac{4 \pi}{3}(8-3 \sqrt{3}). \end{align*}

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