oliadas73

## Answered question

2022-10-02

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:
${\int }_{0}^{2\phantom{\rule{thinmathspace}{0ex}}\pi }{\int }_{0}^{R}{\int }_{0}^{\pi }1\phantom{\rule{thinmathspace}{0ex}}{R}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\phi$.
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
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\phi \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\phi \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)\\ R\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)\end{array}\right)$
I get for the first border:
$\sqrt{{R}^{2}\left(1-{\mathrm{cos}}^{2}\left(\theta \right)-{\mathrm{sin}}^{2}\left(\theta \right)\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sin}}^{2}\left(\phi \right)\right)}$. What's that even?

### Answer & Explanation

selixaak

Beginner2022-10-03Added 6 answers

Step 1
Such as $x=-\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}$
${\int }_{-R}^{R}{\int }_{-\sqrt{{R}^{2}-{z}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}}}{\int }_{-\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}^{\sqrt{{R}^{2}-{z}^{2}-{y}^{2}}}1\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{y}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{z}$
so we can also do
$2{\int }_{0}^{\pi }{\int }_{0}^{2\pi }{\int }_{0}^{R}{\rho }^{2}sen\phi \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\rho \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\phi$
with $x=\rho sen\left(\phi \right)cos\left(\theta \right)$
$y=\rho sen\left(\phi \right)sen\left(\theta \right)$
$z=\rho cos\left(\phi \right)$
Step 2
Then ${x}^{2}+{y}^{2}+{z}^{2}={\rho }^{2}se{n}^{2}\left(\phi \right)\left(co{s}^{2}\left(\theta \right)+se{n}^{2}\left(\theta \right)\right)+{\rho }^{2}co{s}^{2}\left(\phi \right)={\rho }^{2}se{n}^{2}\left(\phi \right)+{\rho }^{2}co{s}^{2}\left(\phi \right)={\rho }^{2}$

Lisantiom

Beginner2022-10-04Added 2 answers

Explanation:
If $x=R\mathrm{cos}\left(\phi \right)\mathrm{sin}\left(\theta \right)$, $y=R\mathrm{sin}\left(\phi \right)\mathrm{sin}\left(\theta \right)$, and $z=R\mathrm{cos}\left(\theta \right)$, then
$\begin{array}{rl}{x}^{2}+{y}^{2}+{z}^{2}& ={R}^{2}{\mathrm{cos}}^{2}\left(\phi \right){\mathrm{sin}}^{2}\left(\theta \right)+{R}^{2}{\mathrm{sin}}^{2}\left(\phi \right){\mathrm{sin}}^{2}\left(\theta \right)+{R}^{2}{\mathrm{cos}}^{2}\left(\theta \right)\\ & ={R}^{2}\left({\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)\right)\\ & ={R}^{2}.\end{array}$

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