Finding volume of sphere cartesian to polar

Using cartesian coordinates the volume of a sphere can be calculated via

${\int}_{-R}^{R}{\int}_{-\sqrt{{R}^{2}-{z}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}}}{\int}_{-\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}1\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{y}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{z}$

However using a parametrisation with polar coordinates the Integral becomes:

${\int}_{0}^{2\phantom{\rule{thinmathspace}{0ex}}\pi}{\int}_{0}^{R}{\int}_{0}^{\pi}1\phantom{\rule{thinmathspace}{0ex}}{R}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\phi $.

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

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\phi )\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\phi )\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )\end{array}\right)$

I get for the first border:

$\sqrt{{R}^{2}(1-{\mathrm{cos}}^{2}(\theta )-{\mathrm{sin}}^{2}(\theta )\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sin}}^{2}(\phi ))}$. What's that even?

Using cartesian coordinates the volume of a sphere can be calculated via

${\int}_{-R}^{R}{\int}_{-\sqrt{{R}^{2}-{z}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}}}{\int}_{-\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}1\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{y}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{z}$

However using a parametrisation with polar coordinates the Integral becomes:

${\int}_{0}^{2\phantom{\rule{thinmathspace}{0ex}}\pi}{\int}_{0}^{R}{\int}_{0}^{\pi}1\phantom{\rule{thinmathspace}{0ex}}{R}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\phi $.

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

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\phi )\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\phi )\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )\end{array}\right)$

I get for the first border:

$\sqrt{{R}^{2}(1-{\mathrm{cos}}^{2}(\theta )-{\mathrm{sin}}^{2}(\theta )\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sin}}^{2}(\phi ))}$. What's that even?