finding volume of sphere cartesian to polar

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?

Solution 1:

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 $$x^2+y^2+z^2=\rho^2sen^2(\varphi)(cos^2(\theta)+sen^2(\theta))+\rho^2cos^2(\varphi)=\rho^2sen^2(\varphi)+\rho^2cos^2(\varphi)=\rho^2$$

Solution 2:

If $x=R\cos(\varphi)\sin(\theta)$, $y=R\sin(\varphi)\sin(\theta)$, and $z=R\cos(\theta)$, then\begin{align}x^2+y^2+z^2&=R^2\cos^2(\varphi)\sin^2(\theta)+R^2\sin^2(\varphi)\sin^2(\theta)+R^2\cos^2(\theta)\\&=R^2\bigl(\sin^2(\theta)+\cos^2(\theta)\bigr)\\&=R^2.\end{align}