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9.3. Polar Coordinates

Figure B. René Descartes (1596-1650).

Philosopher and mathematician René Descartes originated the idea of describing plane geometry using (x, y) coordinates measured from a pair of perpendicular coordinate axes. These rectangular coordinates are known as Cartesian coordinates, in his honor.

Figure C. Polar coordinates.

As a logical extension of Descartes' idea, one can find different ways of defining coordinates on the plane, such as the polar coordinates in Figure C. In polar coordinates, the differential of area, Figure D can be written as da = R dR dφ. The idea is that since dR and dφ are infinitesimally small, the shaded area in the figure is very nearly a rectangle, measuring dR is one dimension and Rdφ in the other. (The latter follows from the definition of radian measure.)

Figure D. The differential of area in polar coordinates.

Example 98

◊ A disk has mass M and radius b. Find its moment of inertia for rotation about the axis passing perpendicularly through its center.

\begin{align} I &= \int R^2 dM \\ &= \int R^2 \frac{dM}{da}da \\ &= \int R^2 \frac{M}{\pi b^2}da \\ &= \frac{M}{\pi b^2}\int_{R=0}^b \int_{\phi=0}^{2\pi} R^2 \cdot R d \phi dR \\ &= \frac{M}{\pi b^2}\int_{R=0}^b R^3 \int_{\phi=0}^{2\pi} d \phi dR \\ &= \frac{2 M}{ b^2}\int_{R=0}^b R^3 dR \\ &= \frac{M b^4}{2} \end{align}

Figure E. The function e-x2, Example 99.

Example 99

In statistics, the standard “bell curve” (also known as the normal distribution or Gaussian) is shaped like e-x2. An area under this curve is proportional to the probability that x lies within a certain range. To fix the constant of proportionality, we need to evaluate

\[ I = \int_{-\infty}^\infty e^{-x^2} dx , \]

which corresponds to a probability of 1. As discussed on p. 95, the corresponding indefinite integral can't be done in closed form. The definite integral from -∞ to +∞, however, can be evaluated by the following devious trick due to Poisson. We first write I2 as a product of two copies of the integral.

\[ I^2 = \left(\int_{-\infty}^\infty e^{-x^2} dx\right) \left(\int_{-\infty}^\infty e^{-x^2} dx\right) \]

Since the variable of integration x is a “dummy” variable, we can choose it to be any letter of the alphabet. Let's change the second one to y:

\[ I^2 = \left(\int_{-\infty}^\infty e^{-x^2} dx\right) \left(\int_{-\infty}^\infty e^{-y^2} dy\right) \]

This is in principle a pointless and trivial change, but it suggests visualizing the right-hand side in the Cartesian plane, and considering it as the integral of a single function that depends on both x and y:

\[ I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty \left( e^{-y^2}e^{-x^2}\right) dx dy \]

Switching to polar coordinates, we have

\begin{align} I^2 &= \int_0^{2\pi} \int_0^\infty e^{-R^2} R dR d \phi \\ &= 2\pi \int_0^\infty e^{-R^2} R dR , \end{align}

which can be done using the substitution u=R2, du = 2R dR:

\begin{align} I^2 &= 2\pi \int_0^\infty e^{-u} (du/2) \\ &= \pi \\ I &= \sqrt{\pi} \end{align}