Editing Polar coordinate system (section)

===Integral calculus (area)===
[[Image:Polar coordinates integration region.svg|thumb|The integration region ''R'' is bounded by the curve ''r''(''φ'') and the rays ''φ'' = ''a'' and ''φ'' = ''b''.]]
Let ''R'' denote the region enclosed by a curve ''r''(''φ'') and the rays ''φ'' = ''a'' and ''φ'' = ''b'', where {{nowrap|0 < ''b'' − ''a'' ≤ 2{{pi}}}}. Then, the area of ''R'' is
<math display="block">\frac12\int_a^b \left[r(\varphi)\right]^2\, d\varphi.</math>

[[Image:Polar coordinates integration Riemann sum.svg|thumb|The region ''R'' is approximated by ''n'' sectors (here, ''n'' = 5).]]
[[File:Planimeter.jpg|thumb|A [[planimeter]], which mechanically computes polar integrals]]
This result can be found as follows. First, the interval {{nowrap|[''a'', ''b'']}} is divided into ''n'' subintervals, where ''n'' is some positive integer. Thus Δ''φ'', the angle measure of each subinterval, is equal to {{math|''b'' − ''a''}} (the total angle measure of the interval), divided by ''n'', the number of subintervals. For each subinterval ''i'' = 1, 2, ..., ''n'', let ''φ''<sub>''i''</sub> be the midpoint of the subinterval, and construct a [[circular sector|sector]] with the center at the pole, radius ''r''(''φ''<sub>''i''</sub>), central angle Δ''φ'' and arc length ''r''(''φ''<sub>''i''</sub>)Δ''φ''. The area of each constructed sector is therefore equal to
<math display="block">\left[r(\varphi_i)\right]^2 \pi \cdot \frac{\Delta \varphi}{2\pi} = \frac{1}{2}\left[r(\varphi_i)\right]^2 \Delta \varphi.</math>
Hence, the total area of all of the sectors is
<math display="block">\sum_{i=1}^n \tfrac12r(\varphi_i)^2\,\Delta\varphi.</math>

As the number of subintervals ''n'' is increased, the approximation of the area improves. Taking {{nowrap|''n'' → ∞}}, the sum becomes the [[Riemann sum]] for the above integral.

A mechanical device that computes area integrals is the [[planimeter]], which measures the area of plane figures by tracing them out: this replicates integration in polar coordinates by adding a joint so that the 2-element [[Linkage (mechanical)|linkage]] effects [[Green's theorem]], converting the quadratic polar integral to a linear integral.

====Generalization====
Using [[Cartesian coordinates]], an infinitesimal area element can be calculated as ''dA'' = ''dx'' ''dy''. The [[integration by substitution#Substitution for multiple variables|substitution rule for multiple integrals]] states that, when using other coordinates, the [[Jacobian matrix and determinant|Jacobian]] determinant of the coordinate conversion formula has to be considered:
<math display="block">J = \det \frac{\partial(x, y)}{\partial(r, \varphi)}
  = \begin{vmatrix}
      \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \varphi} \\[2pt]
      \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \varphi}
    \end{vmatrix}
  = \begin{vmatrix}
      \cos\varphi & -r\sin\varphi \\
      \sin\varphi &  r\cos\varphi
    \end{vmatrix}
  = r\cos^2\varphi + r\sin^2\varphi = r.
</math>

Hence, an area element in polar coordinates can be written as
<math display="block">dA = dx\,dy\ = J\,dr\,d\varphi = r\,dr\,d\varphi.</math>

Now, a function, that is given in polar coordinates, can be integrated as follows:
<math display="block">\iint_R f(x, y)\, dA = \int_a^b \int_0^{r(\varphi)} f(r, \varphi)\,r\,dr\,d\varphi.</math>

Here, ''R'' is the same region as above, namely, the region enclosed by a curve ''r''(''φ'') and the rays ''φ'' = ''a'' and ''φ'' = ''b''. The formula for the area of ''R'' is retrieved by taking ''f'' identically equal to 1.

[[Image:E^(-x^2).svg|thumb|right|A graph of <math>f(x) = e^{-x^2}</math> and the area between the function and the <math>x</math>-axis, which is equal to <math>\sqrt{\pi}</math>.]]

A more surprising application of this result yields the [[Gaussian integral]]:
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt\pi.</math>