# Volume of an Universe with $k=+1$

I read in the Steven Weinberg’s book “Cosmology”:

So far, we have considered only local properties of the spacetime. Now let us look at it in the large. For $k = +1$ space is finite, though like any spherical surface it has no boundary. The coordinate system used to derive Eq. (1.1.7) $$ds^2=a^2 \left [d\boldsymbol{x}^2 + k\frac{(\boldsymbol{x}\cdot d\boldsymbol{x})^2}{1-k\boldsymbol{x}^2}\right ] \tag{1.1.7}$$ with $k = +1$ only covers half the space, with $z > 0$, in the same way that a polar projection map of the earth can show only one hemisphere. Taking account of the fact that z can have either sign, the circumference of the space is $2\pi a$, and its volume is $2 \pi^2 a^3$

I do not understand why Volume is

$$V = 2 \pi^2 a^3$$

How do you demonstrate this mathematical expression?

In spherical coordinates, the spatial part of the metric has the form $$\text{d}\sigma^2 = a^2\left(\frac{\text{d}r^2}{1-kr^2} + r^2\text{d}\theta^2 + r^2\sin^2\theta\text{d}\varphi^2\right).$$ You can derive this using \begin{align} \boldsymbol{x}^2 &= r^2,\\ \text{d}\boldsymbol{x}^2 &= \text{d}r^2 + r^2\text{d}\theta^2 + r^2\sin^2\theta\text{d}\varphi^2,\\ \boldsymbol{x}\cdot \text{d}\boldsymbol{x} &= r\text{d}r. \end{align} If $k=1$, we can use the substitution $r=\sin\chi$ to re-write this as $$\text{d}\sigma^2 = g_{ij}\text{d}x^i \text{d}x^j = a^2\left(\text{d}\chi^2 + \sin^2\chi\text{d}\theta^2 + \sin^2\chi\sin^2\theta\text{d}\varphi^2\right),$$ with $(x^1,x^2,x^3) = (\chi,\theta,\varphi)$. This is the metric of a 3-sphere, expressed in hyperspherical coordinates. The total volume of a 3-sphere is $$V = \int_{S^3}\sqrt{|\det g|}\,\text{d}\chi\text{d}\theta\text{d}\varphi = a^3\int_0^\pi\sin^2\chi\text{d}\chi\int_0^{\pi}\sin\theta\text{d}\theta\int_0^{2\pi}\text{d}\varphi = 2\pi^2a^3.$$
$$S_{n-1} = \frac{2 \pi^{\frac n 2}}{\Gamma \left (\frac n 2 \right )} \, R^{n-1}$$
$$2 \pi^2 a^3$$