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I have a figure C that is defined as the intersection between the sphere $x^2+y^2+z^2 \le 1 $ and the cyllinder $x^2+y^2 \le \frac{1}{4}$.
How should i calculate the volume of this figure?

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    One-variable calculus and the [shell method](https://en.wikipedia.org/wiki/Shell_integration) is probably easiest; this amounts to Yves' calculation (with $\theta$ integrated out). (The [disk method](https://en.wikipedia.org/wiki/Disc_integration) also works, which amounts to Jack's calculation.) – 2017-01-18
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    Do you mean the volume of remaining sphere alone (sphere with a hole in it) or the volume of the cylinder with two spherical caps on its bases? – 2017-01-18
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    it is intriguing how the remaining volume depends on a single parameter $h$ only, as $\pi h^3/6$..what are other such examples? – 2017-01-19

5 Answers 5

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The volume $\ds{V}$ is given by $\ds{\pars{~\mbox{brackets expressions are}\ {\texttt Iverson\ Brackets}~}}$:

\begin{align} V & = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \bracks{x^{2} + y^{2} + z^{2} \leq 1}\bracks{x^{2} + y^{2} \leq {1 \over 4}} \dd x\,\dd y\,\dd z \\[5mm] & = \int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}\bracks{\rho^{2} + z^{2} \leq 1}\bracks{\rho^{2} \leq {1 \over 4}}\, \rho\,\dd\rho\,\dd\theta\,\dd z\label{1}\tag{1} \end{align} In \eqref{1}, we switched to cylindric coordinates. Then, \begin{align} V & = 2\pi\int_{-\infty}^{\infty}\int_{0}^{1/2} \bracks{-\root{1 - \rho^{2}} \leq z \leq \root{1 - \rho^{2}}}\, \rho\,\dd\rho\,\dd z = 4\pi\int_{0}^{1/2}\root{1 - \rho^{2}}\rho\,\dd\rho \\[5mm] & = \bbx{\ds{{8 - 3\root{3} \over 6}\,\pi}} \approx 1.4681 \end{align}

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In cylindrical coordinates,

$$\int_0^{2\pi}2\int_0^{1/2}\sqrt{1-r^2}\,r\,dr\,d\theta=2\cdot2\pi\cdot\frac{\bar1}3\left.(1-r^2)^{3/2}\right|_0^{1/2}=\frac{4\pi}3\left(1-\frac34\sqrt{\frac34}\right).$$

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The given cylinder and the given sphere share the sections given by $z=\pm\frac{\sqrt{3}}{2}$.
If $|z|\leq\frac{\sqrt{3}}{2}$, the area of the section is $\frac{\pi}{4}$. If $\frac{\sqrt{3}}{2}\leq|z|\leq 1$, the area of the section is $\pi(1-z^2)$.
Integrating $1$ on sections, we get that the volume is given by:

$$ \frac{\pi}{4}\sqrt{3}+2\pi\int_{\sqrt{3}/2}^{1}(1-z^2)\,dz =\color{red}{\frac{\pi}{6}(8-3\sqrt{3})}.$$

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Here is a sketch of the sphere and part of the infinite length cylinder:

intersection sphere with cylinder (Large version)

It suggests to split the intersection volume into three volumes:

  • the fully enclosed cylinder part $V_m$ in the middle and
  • the two spherical caps, one on top $V_t$, one at the bottom $V_t$, where by symmetry $V_t = V_b$

At the cap base circles we have both $$ x^2 + y^2 + z^2 = 1 \\ x^2 + y^2 = 1/4 $$ Subtracting gives $$ z^2 = 3/4 \iff \\ z = \pm \sqrt{3}/2 $$ This gives $$ V_m= \pi r^2 h = \pi (1/2)^2 \cdot 2 \cdot \sqrt{3}/2 = \pi \sqrt{3}/4 $$

For the top cap we stack up disc shaped volume elements $$ dV = A(z) \, dz $$ where $A(z)$ is the area of a disc at height $z$ and get \begin{align} V_t &= \int\limits_{\sqrt{3}/2}^1 A(z) \, dz \\ &= \int\limits_{\sqrt{3}/2}^1 \pi r(z)^2 \, dz \\ &= \int\limits_{\sqrt{3}/2}^1 \pi (1-z^2) \, dz \\ &= \pi \, \left[z - z^3/3 \right]_{\sqrt{3}/2}^1 \\ &= \pi \left( 1-1/3 - (\sqrt{3}/2 - (\sqrt{3}/2)^3/3) \right) \\ &= \pi \left( 2/3 - \sqrt{3}/2 + \sqrt{3}/8 \right) \\ &= \pi \left( 2/3 - 3\sqrt{3}/8 \right) \\ \end{align} This gives \begin{align} V &= V_t + V_m + V_b \\ &= \pi \sqrt{3}/4 + 2 \pi \left( 2/3 - 3 \sqrt{3} / 8 \right) \\ &= \pi \left( 4/3 - 2 \sqrt{3}/4 \right) \\ &= \pi \left( 4/3 - \sqrt{3}/2 \right) \end{align}

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    Thank you! But I can't understand the first step you take calculating Vt. – 2017-01-23
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    @jogan It is done by stacking up disc shaped volume elements along the $z$ axis, each disc centered around the origin. – 2017-01-24
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    Why is the radius $(1 - z^2)$ though? – 2017-01-24
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    From $x^2 + y^2 + z^2 = 1$ (point on the sphere) and $x^2 + y^2 = r^2$ (circle in a $x$-$y$-plane, $z=\text{const}$) we get $r^2 + z^2 = 1 \iff r = \pm \sqrt{1-z^2}$, where we use the positive solution as $r \ge 0$. – 2017-01-24
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The volume of a sphere with hole and the volume of a cylinder with two caps on its bases: enter image description here