I'm just starting to read Hartle's Gravity and he gives the following equation for the volume of a 2D sphere of radius $r$ if space was a fixed 3-sphere geometry on page 20. $$V = 4\pi a^3\left(\frac{1}{2}\sin^{-1}\left(\frac{r}{a}\right)-\frac{r}{2a}\left[1-\left(\frac{r}{a}\right)^2\right]^{1/2}\right).$$ He then claims that if we were to assume that $r/a$ were small, i.e., the radius of curvature of space $a$ were much smaller than the radius of the 2D sphere, then we would get the normal Euclidean formula for the volume of the sphere. $$V=\frac{4}{3}\pi r^3$$ The inverse sine term goes to 0 if we assume that the argument is small but I'm unsure about how to deal with the square root term. A series expansion does not give me the necessary terms. Any help would be appreciated!
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1$\begingroup$ $sin(x)\approx x$ for very small $x$ now think of arcsin function. Also you will need to calculate the expansion to third order since first order will cancel out there will be another term from arcsin expansion to get the right numerical factor of $r^3$ $\endgroup$– aitfelCommented Jan 16, 2022 at 9:12
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The arcsine function terms and the root formula terms are expanded into series to the terms with different coefficients, and then computed. $$\frac{1}{2}\sin^{-1}\left(\frac{r}{a}\right)=\frac{r}{2a}+\frac{1}{12}\frac{r^3}{a^3}+o\left(r^3\right)$$ $$-\frac{r}{2a}\left[1-\left(\frac{r}{a}\right)^2\right]^{1/2}=-\frac{r}{2a}+\frac{r^3}{4a^3}+o\left(r^3\right)$$ \begin{align*} V &= 4\pi a^3\left(\frac{r}{2a}+\frac{1}{12}\frac{r^3}{a^3}+o\left(r^3\right)-\frac{r}{2a}+\frac{r^3}{4a^3}+o\left(r^3\right)\right)\\ &=\frac{4}{3}\pi r^3+o\left(r^3\right). \end{align*}
