Calculus/Polar Integration

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Contents

[edit] Introduction

Integrating a polar equation requires a different approach than integration under the Cartesian system, hence yielding a different formula, which is not as straightforward as integrating the function f(x).

[edit] Proof

In creating the concept of integration, we used Riemann sums on rectangles to approximate the area under the curve. However, with polar graphs, one must use triangles that start from the origin, and have a radius ending on the curve. If you don't mind skipping the proof, this is the form to use to integrate a polar expression of the form r = f(θ):

\frac{1}{2} \int_{a}^{b} r^2\,d\theta,

where (a,f(a)) and (b,f(b)) are the ends of the curve that you wish to integrate.

[edit] Integral calculus

The integration region R is bounded by the curve r = f(θ) and the rays θ = a and θ = b.

Let R denote the region enclosed by a curve r = f(θ) and the rays θ = a and θ = b, where 0 < b − a < 2π. Then, the area of R is

\frac12\int_a^b r^2\,d\theta.
The region R is approximated by n sectors (here, n = 5).

This result can be found as follows. First, the interval [a,b] is divided into n subintervals, where n is an arbitrary positive integer. Thus θ, the length of each subinterval, is equal to b − a (the total length of the interval), divided by n, the number of subintervals. For each subinterval i=1,2,\ldots,n, let θi be the midpoint of the subinterval, and construct a circular sector with the center at the origin, radius ri = f(θi), central angle δθ, and arc length riδθ. The area of each constructed sector is therefore equal to \tfrac12 r_i^2\delta\theta. Hence, the total area of all of the sectors is

\sum_{i=1}^n \tfrac12 r_i^2\,\delta\theta.

As the number of subintervals n is increased, the approximation of the area continues to improve. In the limit as n\to\infty, the sum becomes the Riemann integral.

[edit] Generalization

Using Cartesian coordinates, an infinitesimal area element can be calculated as dA = dx\,dy. The substitution rule for multiple integrals states that, when using other coordinates, the Jacobian determinant of the coordinate conversion formula has to be considered:

J = \det\frac{\partial(x,y)}{\partial(r,\theta)} =\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} =\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} =r\cos^2\theta + r\sin^2\theta = r.

Hence, an area element in polar coordinates can be written as

dA = J\,dr\,d\theta = r\,dr\,d\theta.

Now, a function that is given in polar coordinates can be integrated as follows:

\iint_R g(r,\theta) \, dA = \int_a^b \int_0^{r(\theta)} g(r,\theta)\,r\,dr\,d\theta.

Here, R is the same region as above, namely, the region enclosed by a curve r = f(θ) and the rays θ = a and θ = b.

The formula for the area of R mentioned above is retrieved by taking g identically equal to 1.

[edit] Applications

Polar integration is often useful when the corresponding integral is either difficult or impossible to do with the Cartesian coordinates. For example, let's try to find the area of the closed unit circle. That is, the area of the region enclosed by x2 + y2 = 1.

[edit] In Cartesian

\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \, dy \, dx = 2\int_{-1}^1 \sqrt{1-x^2} \, dx

In order to evaluate this, one usually uses trigonometric substitution. By setting sinθ = x, we get both \cos\theta = \sqrt{1-x^2} and \cos\theta\,d\theta = dx.

\begin{align}\int\sqrt{1-x^2}\,dx &= \int \cos^2\theta\,d\theta\\ &= \int \frac{1}{2} + \frac{1}{2} \cos 2\theta\,d\theta\\ &= \frac{\theta}{2} + \frac{1}{4}\sin2\theta+c= \frac{\theta}{2} + \frac{1}{2}\sin\theta\cos\theta+c\\ &= \frac{\arcsin x}{2}+\frac{x\sqrt{1-x^2}}{2}+c\end{align}

Putting this back into the equation, we get

2\int_{-1}^1\sqrt{1-x^2}\,dx = 2\left[\frac{\arcsin x}{2}+\frac{x\sqrt{1-x^2}}{2}\right]_{-1}^1 = \arcsin 1-\arcsin(-1) = \pi

[edit] In Polar

To integrate in polar coordinates, we first realize r = \sqrt{x^2 + y^2} = \sqrt{1} = 1 and in order to include the whole circle, a = 0 and b = 2π.

\int_{0}^{2\pi} \int_{0}^1 r\,dr\,d\theta = \int_0^{2\pi} \left[\frac{r^2}{2}\right]_0^1\,d\theta = \int_0^{2\pi} \frac{1}{2}\,d\theta = \left[\frac{\theta}{2}\right]_0^{2\pi} = \frac{2\pi}{2} = \pi

[edit] An interesting example

A less intuitive application of polar integration yields the Gaussian integral

\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt\pi.

Try it! (Hint: multiply \int_{-\infty}^\infty e^{-x^2} \, dx and \int_{-\infty}^\infty e^{-y^2} \, dy.)

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