Using cartesian coordinates the volume of a sphere can be calculated via
$$\int_{-R}^{R}\int_{-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}}\int_{-\sqrt{R^2-z^2-y^2}}^{\sqrt{R^2-z^2-y^2}}1\,\mathrm{dx\,dy\,dz}$$
However using a parametrisation with polar coordinates the Integral becomes:
$$\int_{0}^{2\,\pi}\int_{0}^{R}\int_{0}^{\pi} 1\,R^2\,\sin(\theta)\,\mathrm{d\theta}\,\mathrm{dR\,d\varphi}$$
Thinking closer about this I just don't see how these integrals are interchangeable. Notably substituting polar coordinates in the cartesian one I don't notice how the borders simplify.
For instance using the substitution $\left(\begin{array}{c}x \\ y \\z\end{array}\right)=\left(\begin{array}{c}R\,\cos(\varphi)\,\sin(\theta) \\ R\,\sin(\varphi)\,\sin(\theta)\\ R\,\cos(\theta)\end{array}\right)$ I get for the first border:
$\sqrt{R^2(1-\cos^2(\theta)-\sin^2(\theta)\,\sin^2(\varphi))}$. What's that even?
Best Answer
Such as $x=-\sqrt{R^2-z^2-y^2}$ to $\sqrt{R^2-z^2-y^2}$ $$\int_{-R}^{R}\int_{-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}}\int_{-\sqrt{R ^2-z^2-y^2}}^{\sqrt{R^2-z^2-y^2}}1\,\mathrm{dx\,dy\,dz}$$
so we can also do
$$2\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho^2sen\varphi \,\mathrm{d\rho\,d\theta\,d\varphi}$$
with
$$x=\rho sen(\varphi)cos(\theta)$$ $$y=\rho sen(\varphi)sen(\theta)$$ $$z=\rho cos(\varphi)$$
then $$x^2+y^2+z^2=\rho^2sen^2(\varphi)(cos^2(\theta)+sen^2(\theta))+\rho^2cos^2(\varphi)=\rho^2sen^2(\varphi)+\rho^2cos^2(\varphi)=\rho^2$$