Integration using polar coordinates

Integration using polar coordinates is a technique for solving integrals using polar coordinates. Sometimes an integral that is complicated in one set or coordinates, such as Cartesian coordinates become very easy or even trivial in polar coordinates.

Example

Consider the integral

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \int^{\infty}_{-\infty} e^{-x^2} \, \text{d}x}

This can be converted into polar coordinated by multiplying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} by itself, so that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I^2 = \int^{\infty}_{-\infty} e^{-x^2} \, \text{d}x \int^{\infty}_{-\infty} e^{-y^2} \, \text{d}y}

This can be expressed as the double integral,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I^2 = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-(x^2 + y^2)} \, \text{d}x \, \text{d}x}

Now the usefulness of polar coordinates becomes apparent as in polar coordinates, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r^2 = x^2 + y^2} . The bounds mean the integral is over the entire x-y plane, so $r$ varies from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{0}} to $\infty$ and $\phi$ from ${\text{0}}$ to $2\pi$. To convert the differentials, we must multiply by the Jacobian, $r$ to get

$I^{2}=\int _{r=0}^{r=\infty }\int _{\phi =0}^{\phi =2\pi }re^{-r^{2}}\,{\text{d}}r\,{\text{d}}\phi$

This can be integrated by separating the integral. The $\phi$ integral gives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi} and the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} integral gives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{1}} . This means the initial integral, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2 \pi}} .