# What is wrong with my reasoning regarding finding volumes by integration?

The problem from the book is (this is Calculus 2 stuff):

Find the volume common to two spheres, each with radius $r$, if the center of each sphere lies on the surface of the other sphere.

I put the center of one sphere at the origin, so its equation is $x^2 + y^2 + z^2 = r^2$. I put the center of the other sphere on the $x$-axis at $r$, so its equation is $(x-r)^2 + y^2 + z^2 = r^2$.

By looking at the solid down the $y$- or $z$-axis it looks like a football. By looking at it down the $x$-axis, it looks like a circle. So, the spheres meet along a plane as can be confirmed by setting the two equations equal to each other and simplifying until you get $x = r/2$.

So, my strategy is to integrate down the $x$-axis from 0 to $r/2$, getting the volume of the cap of one of the spheres and just doubling it, since the solid is symmetric. In other words, I want to take circular cross-sections along the $x$-axis, use the formula for the area of a circle to find their areas, and add them all up.

The problem with this is that I need to find an equation for $r$ in terms of $x$, and it has to be quadratic rather than linear, otherwise I'll end up with the volume of a cone rather than a sphere. But when I solve for, say, $y^2$ in one equation, plug it into the other one, and solve for $r$, I get something like $r = \sqrt{2 x^2}$, which is linear.

-

If you restrict yourself to the $xy$ plane you have two intersecting circles. Now you want $y$ as a function of $x$ for the circle centered at $x=r$, so you can solve the equation for $y$, getting $y=\sqrt{r^2-x^2}$. Your confusion may have come from reusing $r$, once for the radius of the spheres and once for the radius perpendicular to the $x$ axis out to the right hand sphere.

-
Oh, I see it now! I am reusing r. Thank you! – Matt Gregory Oct 3 '11 at 17:26

The analytic geometry of $3$-dimensional space is not needed to solve this problem. In particular, there is no need for the equations of the spheres. All we need is some information about the volumes of solids of revolution.

Draw two circles of radius $1$, one with center $(0,0)$, the other with center $(1,0)$. (Later we can scale everything by the linear factor $r$, which scales volume by the factor $r^3$.)

The volume we are looking for is twice the volume obtained by rotating the part of the circle $x^2+y^2=1$, from $x=1/2$ to $x=1$, about the $x$-axis. (This is clear if we have drawn a picture). So the desired volume, in the case $r=1$, is $$2\int_{1/2}^1 \pi y^2\,dx.$$ There remains some work to do, but it should be straightforward.

Comment: There are other approaches. Volumes of spherical caps were known well before the formal discovery of the calculus. And if we do use calculus, we can also tackle the problem by making perpendicular to the $y$-axis, or by using the "shell" method. It is worthwhile experimenting with at least one of these methods. The algebra gets a little more complicated.

-
Yeah, I know 3D stuff is not needed, but I was just trying to figure out why it goes wrong because I keep going down the same dead end over and over on these types of problems. I figured if I knew why it was wrong, more complicated, whatever, then I would quit going down that road. Thanks for the reply! – Matt Gregory Oct 3 '11 at 17:57
Calculus of one variable is (usually) a lot easier than calculus of several variables, or at least much more familiar. So if we can solve a volume problem by slicing, or by shells, that is likely to be easier. Note that the $r/2$ that you got by calculation is clear if you draw the two circles and use symmetry. – André Nicolas Oct 3 '11 at 18:43