Find volume of remaining solid using surfaces of revolution. A sphere has radius R. A cylindrical hole has been drilled straight through the center of the sphere. What's the volume of the remaining solid if the height of the remaining solid is 6 cm high?
 A: Great problem.  The result is surprising.
First, place the sphere's center at the origin, and look at it from the perspective of the positive $z$-axis.
Now, notice that the equation of the circle of radius $R$ that you see is $x^2+y^2=R^2$.  This shows that the radius of a horizontal cross section of the sphere is $x=\sqrt{R^2-y^2}$.  The radius of the cylinder can be found using Pythagorean's theorem.  It is $\sqrt{R^2-9}$.  It follows that the area of a horizontal cross section of the shape is:
$$A(y)=\pi(R^2-y^2)-\pi(R^2-9)=\pi(9-y^2)$$
If we integrate this from $y=-3$ to $y=3$ we find the volume of the shape to be $36\pi$.  Note this answer doesn't depend on the radius of the sphere! (Although I guess that we assume $R\ge 3$ to stay in the world of real numbers).
You can check that our final answer makes sense in the limiting case that $R=3$.  In that case, there is no whole, and the volume of a sphere of radius $3$ is indeed $36\pi$.
A: We imagine the same thing done to a half-sphere. Then we can multiply by $2$ at the end. 
Let the radius of the cylinder be $r$. Then by the Pythagorean Theorem, $r^2=R^2-9$. 
If we take a horizontal slice at height $x$, the cross-section is a circle of radius $\sqrt{R^2-x^2}$, with a hole of radius $r$. Thus the cross-sectional area is $\pi(R^2-x^2 -r^2)$. This is $9-x^2$. Integrate from $x=0$ to $x=3$. Then double to get the full remaining volume.
Note that the answer depends neither on $r$ nor $R$! 
