Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having trouble expressing the titular question as iterated integrals over a given region. I have tried narrowing down the problem, and have concluded that the simplest way to approach this is to integrate over the XZ plane in the positive octant and multiply by 8, but I am having trouble identifying the bounding functions.

share|cite|improve this question
+1 for saying what you've tried. – J. M. Nov 4 '10 at 11:30
up vote 16 down vote accepted

The solid lies above the region $D$ in the $xy$-plane bounded by the circle $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$

Therefore the required volume of the solid is: $$\int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} 2\sqrt{r^{2}-y^{2}} \ dx dy = \frac{16}{3}r^{3}$$

share|cite|improve this answer
Perfect! Short. Succint. Beautiful : ) Would have given +1 had I had the rep! – Kris Nov 4 '10 at 11:54

This is one of those results in calculus which were anticipated by Archimedes. He gave a correct formula for the volume but it is not known exactly how Archimedes solved this problem. There is, however, a simple way to obtain the answer without much calculus. Let me quote from late Gardner's The Unexpected Hanging and Other Mathematical Diversions (Gardner considers the case $r=1$ but this is not essential, of course):

Imagine a sphere of unit radius inside the volume common to the two cylinders and having as its center the point where the axes of the cylinders intersect. Suppose that the cylinders and sphere are sliced in half by a plane through the sphere's center and both axes of the cylinders. The cross section of the volume common to the cylinders will be a square. The cross section of the sphere will be a circle that fills the square.

alt text alt text

Now suppose that the cylinders and sphere are sliced by a plane that is parallel to the previous one but that shaves off only a small portion of each cylinder (have a look at the picture on the left). This will produce parallel tracks on each cylinder, which intersect as before to form a square cross section of the volume common to both cylinders. Also as before, the cross section of the sphere will be a circle inside the square. It is not hard to see (with a little imagination and pencil doodling) that any plane section through the cylinders, parallel to the cylinders' axes, will always have the same result: a square cross section of the volume common to the cylinders, enclosing a circular cross section of the sphere. Think of all these plane sections as being packed together like the leaves of a book. Clearly, the volume of the sphere will be the sum of all circular cross sections, and the volume of the solid common to both cylinders will be the sum of all the square cross sections. We conclude, therefore, that the ratio of the volume of the sphere to the volume of the solid common to the cylinders is the same as the ratio of the area of a circle to the area of a circumscribed square. A brief calculation shows that the latter ratio is $\pi/4$. This allows the following equation, in which $x$ is the volume we seek:

$$\frac{4\pi r^3/3}{x}=\frac{\pi}{4}.$$

The $\pi$'s drop out, giving $x$ a value of $16r^3/3$. The radius in this case is 1, so the volume common to both cylinders is $16/3$. As Archimedes pointed out, it is exactly $2/3$ the volume of a cube that encloses the sphere; that is, a cube with an edge equal to the diameter of each cylinder.

share|cite|improve this answer
+1 because Gardner is, and forever will be, awesome. – J. M. Nov 4 '10 at 14:08
That's a very pretty argument! – Hans Lundmark Nov 4 '10 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.