oliadas73

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}(\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?

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 $x=\rho sen(\phi )cos(\theta )$

$y=\rho sen(\phi )sen(\theta )$

$z=\rho cos(\phi )$

Step 2

Then ${x}^{2}+{y}^{2}+{z}^{2}={\rho}^{2}se{n}^{2}(\phi )(co{s}^{2}(\theta )+se{n}^{2}(\theta ))+{\rho}^{2}co{s}^{2}(\phi )={\rho}^{2}se{n}^{2}(\phi )+{\rho}^{2}co{s}^{2}(\phi )={\rho}^{2}$

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 $x=\rho sen(\phi )cos(\theta )$

$y=\rho sen(\phi )sen(\theta )$

$z=\rho cos(\phi )$

Step 2

Then ${x}^{2}+{y}^{2}+{z}^{2}={\rho}^{2}se{n}^{2}(\phi )(co{s}^{2}(\theta )+se{n}^{2}(\theta ))+{\rho}^{2}co{s}^{2}(\phi )={\rho}^{2}se{n}^{2}(\phi )+{\rho}^{2}co{s}^{2}(\phi )={\rho}^{2}$

Lisantiom

Beginner2022-10-04Added 2 answers

Explanation:

If $x=R\mathrm{cos}(\phi )\mathrm{sin}(\theta )$, $y=R\mathrm{sin}(\phi )\mathrm{sin}(\theta )$, and $z=R\mathrm{cos}(\theta )$, then

$\begin{array}{rl}{x}^{2}+{y}^{2}+{z}^{2}& ={R}^{2}{\mathrm{cos}}^{2}(\phi ){\mathrm{sin}}^{2}(\theta )+{R}^{2}{\mathrm{sin}}^{2}(\phi ){\mathrm{sin}}^{2}(\theta )+{R}^{2}{\mathrm{cos}}^{2}(\theta )\\ & ={R}^{2}{\textstyle (}{\mathrm{sin}}^{2}(\theta )+{\mathrm{cos}}^{2}(\theta ){\textstyle )}\\ & ={R}^{2}.\end{array}$

If $x=R\mathrm{cos}(\phi )\mathrm{sin}(\theta )$, $y=R\mathrm{sin}(\phi )\mathrm{sin}(\theta )$, and $z=R\mathrm{cos}(\theta )$, then

$\begin{array}{rl}{x}^{2}+{y}^{2}+{z}^{2}& ={R}^{2}{\mathrm{cos}}^{2}(\phi ){\mathrm{sin}}^{2}(\theta )+{R}^{2}{\mathrm{sin}}^{2}(\phi ){\mathrm{sin}}^{2}(\theta )+{R}^{2}{\mathrm{cos}}^{2}(\theta )\\ & ={R}^{2}{\textstyle (}{\mathrm{sin}}^{2}(\theta )+{\mathrm{cos}}^{2}(\theta ){\textstyle )}\\ & ={R}^{2}.\end{array}$

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