# 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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