Quick Calculation 3.5

A First Course in String Theory

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Show that

\displaystyle{     \begin{aligned}           \text{vol}(B^d)   &= \frac{\pi^{\frac{d}{2}}}{\Gamma \left( 1 + \frac{d}{2} \right)}   \\     \end{aligned} }

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Geometric proof

The relations \displaystyle{V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)} and \displaystyle{A_{n+1}(R)=(2\pi R)V_{n}(R)} and thus the volumes of n-balls and areas of n-spheres can also be derived geometrically. As noted above, because a ball of radius \displaystyle{R} is obtained from a unit ball \displaystyle{B_{n}} by rescaling all directions in \displaystyle{R} times, \displaystyle V_{n}(R) is proportional to \displaystyle{R^{n}}, which implies \displaystyle{{\frac {dV_{n}(R)}{dR}}={\frac {n}{R}}V_{n}(R)}.

Also, \displaystyle{A_{n-1}(R)={\frac {dV_{n}(R)}{dR}}} because a ball is a union of concentric spheres and increasing radius by \displaystyle{\epsilon} corresponds to a shell of thickness \displaystyle{\epsilon}. Thus, \displaystyle{V_{n}(R)={\frac {R}{n}}A_{n-1}(R)}; equivalently, \displaystyle{V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)}.

— Wikipedia on Volume of an n-ball

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\displaystyle{  \begin{aligned}    V &= \int_0^R S dr \\    V(B^d) &= \int_0^R V(S^{d-1}) dR \\     &= \frac{2 \pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})} \int_0^R r^{d-1} dr \\     &= \frac{2 \pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})} \frac{1}{d} R^{d} \\     \end{aligned}}

— Me@2022-05-18 09:08:11 AM

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2022.05.19 Thursday (c) All rights reserved by ACHK