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Given a solid sphere of radius R, remove a cylinder whose central axis goes through the center of the sphere. Let h denote the height of the remaining solid. Calculate the volume of the remaining solid.

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The curious thing you will discover, after the smoke clears, is that the answer is independent of $R$! Apart from that, it is a standard solid of revolution problem. – André Nicolas Feb 16 '12 at 6:55
It is considered impolite to post in the form "Calculate..." rather than asking a question. Moreover, some hints on what you have tried already would be appropriate. – Dirk Feb 16 '12 at 7:11
Does this help ? – Inquest Feb 16 '12 at 7:17

There's a Wikipedia article about this problem:

It is solved neatly via Cavalieri's principle.

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

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As the volume does not depend on the radius, imagine an cylinder with a radius negligible. The volume will be the volume of a sphere with radius h/2.

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The thing is, though, how do you know the volume does not depend on the radius? You don't know it in advance. – Rahul Nov 2 '12 at 22:00

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: $$ \eqalign{ \int_{-h/2}^{h/2} \pi\bigl(\, {\textstyle{h^2\over 4}}-y^2 \,\bigr) \, dy &=2\int_{0}^{h/2} \pi\bigl(\, {\textstyle{h^2\over 4}}-y^2 \,\bigr) \, dy\cr &=2\pi\bigl[ {\textstyle{h^2\over 4}}y-{\textstyle{y^3\over 3} } \bigr]\bigr|_0^{h/2}\cr &= 2\pi\bigl[ {\textstyle{h^3\over 8 }}-{\textstyle{h^3\over 24} } \bigr]\cr &={\pi h^3\over 6}. } $$

enter image description here

(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. )

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What software did you use for this illustration? – Michael Hardy Nov 4 '12 at 16:48
@MichaelHardy I used the javascript library JSXgraph. – David Mitra Nov 4 '12 at 17:33

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!

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Last line was probably meant to be " We end up with $\dfrac{\pi h^3}{6}.$" – Henry Feb 16 '12 at 8:55
@Henry: Thanks, indeed it was. – André Nicolas Feb 16 '12 at 14:48

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