Given a solid sphere of radius R, remove a cylinder whose central axis goes through the center of the sphere.

Solution 1:

It takes me a long time to draw a picture with software, so you will have to do it for me. Without your picture, the solution below will have no meaning.
Please draw the top half of the circle with center the origin and radius $R$. This circle has equation $x^2+y^2=R^2$. I am a little allergic to fractions, so temporarily let $h=2k$.

What does our sphere with a hole in it look like? Draw the vertical line that goes straight up from the point $A(-k,0)$ until it meets the circle at a point we will call $P$. Make $k$ fairly large, say at least $\frac{3}{4}R$. That helps with the visualization later. Also draw the vertical line that goes straight up from $B(k,0)$ until it meets the circle at a point we will call $Q$. Join the two points $P$ and $Q$ by a horizontal line.

Now for the hard part! Let $\mathcal{W}$ be the region above the line $PQ$ but below the curve $y=\sqrt{R^2-x^2}$. The hole was drilled horizontally, along the $x$-axis. All that's left of the original sphere is the solid that we obtain by rotating the region $\mathcal{W}$ about the $x$-axis. This solid is sometimes called a napkin ring. Note that this solid has height $2k$. The radius of the hole is the length of the line segment $AP$. So this radius is $\sqrt{R^2-k^2}$. It is kind of funny to talk about height, since this "height," and the drilling, is along the $x$-direction. Too late to fix.

We first find the volume obtained by rotating the region below $y=\sqrt{R^2-x^2}$, above the $x$-axis, from $x=-k$ to $x=k$. It is standard solid of revolution stuff that the volume is $$\int_{-k}^k \pi (R^2-x^2)\,dx.$$ Evaluate. It is easier to integrate from $0$ to $k$ and double. We get $$2\pi R^2k -\frac{2\pi k^3}{3}.\qquad\qquad(\ast)$$

The hole is simply a cylinder of height $2k$, and radius $AP$, which is $\sqrt{R^2-k^2}$. So integration is unnecessary. The volume of the hole $$\pi(R^2-k^2)(2k).\qquad(\ast\ast) $$

To find the volume of what's left, subtract $(\ast\ast)$ from $(\ast)$. The $\pi R^2 k$ terms cancel, and after some algebra we get $\dfrac{4}{3}\pi k^3$.
Recall that $k=\frac{h}{2}$ and substitute. We end up with $$\frac{\pi h^3}{6}.$$ Note that the answer turned out to be independent of the radius $R$ of the sphere!

Solution 2:

There's a Wikipedia article about this problem: http://en.wikipedia.org/wiki/Napkin_ring_problem

It is solved neatly via Cavalieri's principle.

(Both of these Wikipedia articles were initially created by me.)

Solution 3:

The "napkin ring" is generated by revolving the light pink region $\color{pink}{Q}$, shown below, about the $y$-axis. This volume can be found using the washer method. A typical washer, generated by revolving the line segment $\overline{\color{maroon}{p_yq_y}}$ about the $y$-axis, is shown below. The coordinates of the endpoints of the line segment can be found using the Pythagorean Theorem and the fact that the sphere has radius $R$ and the cylinder has height $h$ (so the top of the cylinder crosses the $y$ axis at $y=h/2$).

The volume is $$ \int_{-h/2}^{h/2} \pi\Bigl[ \, \Bigl(\underbrace{\sqrt{R^2-y^2}}_{\text{outer radius}\atop \text{of the washer} }\ \Bigr)^2-\Bigl(\underbrace{\sqrt{R^2-(h^2/4)}}_{\text{inner radius}\atop \text{of the washer} } \ \ \Bigr)^2\, \Bigr ] \, dy= \int_{-h/2}^{h/2} \pi\bigl(\, {\textstyle{h^2\over 4}}-y^2 \,\bigr) \, dy. $$

Note that, as André Nicolas points out, the volume is independent of $R$.

Calculating the integral above, and taking advantage of symmetry, the volume of the solid is: $$ \eqalign{ \int_{-h/2}^{h/2} \pi\bigl(\, {\textstyle{h^2\over 4}}-y^2 \,\bigr) \, dy &=2\int_{0}^{h/2} \pi\bigl(\, {\textstyle{h^2\over 4}}-y^2 \,\bigr) \, dy\cr &=2\pi\bigl[ {\textstyle{h^2\over 4}}y-{\textstyle{y^3\over 3} } \bigr]\bigr|_0^{h/2}\cr &= 2\pi\bigl[ {\textstyle{h^3\over 8 }}-{\textstyle{h^3\over 24} } \bigr]\cr &={\pi h^3\over 6}.

} $$

(It's probably best to read André's more detailed answer and just refer to the diagram if need be. The labeling and orientation of the diagram, though, is different from André's answer (in particular the $x$ and $y$ axes are interchanged). His region $\cal W$ is the region $\color{pink}{Q}$ above. )