Answer
$V=\frac{\pi^2r^4}{2}$
Work Step by Step
We can find the volume of a 4-dimensional hypersphere with radius R much like how we would find the volume of a 3-dimensional sphere or the area of a circle using multiple integrals.
The volume can be represented by the quadruple integral:
$$V=\iiiint_{R}dV$$
$R$ represents the interior of a 4-dimensional hypersphere centered at the origin with radius $r$.
We can rewrite the integral as an iterated integral with bounds as follows:
$$V=\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\int_{-\sqrt{r^2-x^2-y^2}}^{\sqrt{r^2-x^2-y^2}}\int_{-\sqrt{r^2-x^2-y^2-z^2}}^{\sqrt{r^2-x^2-y^2-z^2}}dw\,dz\,dy\,dx$$
By using symmetry, we can simplify the bounds of the integrals and rewrite the expression as:
$$V=16\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}\int_{0}^{\sqrt{r^2-x^2-y^2}}\int_{0}^{\sqrt{r^2-x^2-y^2-z^2}}dw\,dz\,dy\,dx$$
Solving the innermost integral, we get:
$$V=16\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}\int_{0}^{\sqrt{r^2-x^2-y^2}}{\sqrt{r^2-x^2-y^2-z^2}}\,dz\,dy\,dx$$
Using the substitution $z={\sqrt{r^2-x^2-y^2}}\sin\theta$, changing the bounds, and simplifying the expression, we get:
$$V=16\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}\int_{0}^{\pi/2}(r^2-x^2-y^2)\cos^2\theta\,d\theta\,dy\,dx$$
We can rearrange the integral and simplify the expression a bit more as follows:
$$V=16\int_{0}^{\pi/2}\cos^2\theta\,d\theta\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}(r^2-x^2-y^2)\,dy\,dx\\
=4\pi\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}(r^2-x^2-y^2)\,dy\,dx$$
Solving the innermost integral, we get:
$$V=4\pi\int_{0}^{r}\bigg[r^2y-x^2y-\frac{y^3}{3}\bigg]_{0}^{\sqrt{r^2-x^2}}dx\\
=4\pi\int_{0}^{r}\bigg(r^2\sqrt{r^2-x^2}-x^2\sqrt{r^2-x^2}-\frac{\sqrt{r^2-x^2}^3}{3}\bigg)dx$$
We can now use one final substitution $x=r\sin\phi$ and after changing the bounds and simplifying, we get:
$$V=4\pi\int_0^{\pi/2}\bigg(r^3\cos\phi-r^3\sin^2\phi\cos\phi-\frac{r^3\cos^3\phi}{3}\bigg)r\cos\phi\,d\phi$$
We can simplify, factor out a $r^4$ term from the whole expression and break apart the integral.
$$V=4\pi r^4\bigg(\int_0^{\pi/2}\cos^2\phi-\int_0^{\pi/2}\sin^2\phi\cos^2\phi-\int_0^{\pi/2}\frac{\cos^4\phi}{3}\bigg)d\phi$$
Solving these integrals, we finally arrive at our volume:
$$V=4\pi r^4\bigg(\frac{\pi}{4}-\frac{\pi}{16}-\frac{\pi}{16}\bigg)\\
=\frac{\pi^2r^4}{2}$$
