Proof for volume of n-ball with radius 1

Solution 1:

You have two errors. The formula for volume is wrong by a factor of 2. The volume is proportional to $r^n$, not $r^{n/2}$ $$\text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2}))$$ The second error is that if you want to calculate the volume of the 3D sphere, the $n$ on the right hand side is $2$, not $3$. So $$\text{vol}(B_{3}) = 2 \int_{0}^1 \text{vol}(B_{2}) \cdot (1-x^2)^{2/2} \,\,dx=2\pi\frac23$$

Solution 2:

The correct recurrence is $\text{Vol}(B_1)=\text{length}((-1,1))=2$, and for any $n\geq 1$, \begin{align} \text{Vol}(B_{n+1})&=\int_{-1}^1\text{Vol}(B_n\,; \text{radius} \sqrt{1-x^2})\,dx\\ &=2\int_0^1\text{Vol}(B_n) \left(\sqrt{1-x^2}\right)^n\,dx\\ &=2\text{Vol}(B_n)\int_0^1(1-x^2)^{n/2}\,dx \end{align}

If you wish to evaluate this more explicitly, make the substitution $t=x^2$, then \begin{align} 2\int_0^1(1-x^2)^{n/2}\,dx&=2\int_0^1(1-t)^{n/2}\frac{dt}{2\sqrt{t}}\\ &=\int_0^1t^{-1/2}(1-t)^{n/2}\,dt\\ &:=B\left(\frac{1}{2},\frac{n}{2}+1\right), \end{align} where $B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,dt$ is the Beta function. Using the functional equation, we have that $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, and thus $B\left(\frac{1}{2},\frac{n}{2}+1\right)=\frac{\sqrt{\pi}\,\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\frac{n+1}{2}+1\right)}$, where we used the fact that $\Gamma(\frac{1}{2})=\sqrt{\pi}$. Therefore, \begin{align} \text{Vol}(B_{n+1})&=\text{Vol}(B_n)\cdot \frac{\sqrt{\pi}\,\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\frac{n+1}{2}+1\right)}. \end{align} Now, you can solve this recurrence.