Calculus/Polar Integration

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 $f(x)$ .

Proof

In creating the concept of integration, we used Riemann sums of rectangles to approximate the area under the curve. However, with polar graphs, one can use sectors of circles with radius $r$ and angle measure $d\theta$ . The area of each sector is then $\pi r^{2}{\frac {d\theta }{2\pi }}$ and the sum of all the infinitesimally small sectors' areas is: ${\frac {1}{2}}\int \limits _{a}^{b}r^{2}d\theta$ , This is the form to use to integrate a polar expression of the form $r=f(\theta )$ where $(a,f(a))$ and $(b,f(b))$ are the ends of the curve that you wish to integrate.

Integral calculus

The integration region

R

is bounded by the curve

r=f(\theta )

and the rays

\theta =a

and

\theta =b

.

Let $R$ denote the region enclosed by a curve $r=f(\theta )$ and the rays $\theta =a$ and $\theta =b$ , where $0<b-a<2\pi$ . Then, the area of $R$ is

{\frac {1}{2}}\int \limits _{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 $\theta$ , 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 $i=1,2,\dots ,n$ , let $\theta _{i}$ be the midpoint of the subinterval, and construct a circular sector with the center at the origin, radius $r_{i}=f(\theta _{i})$ , central angle $\delta \theta$ , and arc length $r_{i}\delta \theta$ . The area of each constructed sector is therefore equal to ${\tfrac {1}{2}}r_{i}^{2}\delta \theta$ . Hence, the total area of all of the sectors is

\sum _{i=1}^{n}{\tfrac {1}{2}}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.

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}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}\\{\dfrac {\partial y}{\partial r}}&{\dfrac {\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 \limits _{a}^{b}\int \limits _{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 $r=f(\theta )$ and the rays $\theta =a$ and $\theta =b$ .

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

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 $x^{2}+y^{2}=1$ .

In Cartesian

Template:Organize section

\int \limits _{-1}^{1}\int \limits _{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}\,dy\,dx=2\int \limits _{-1}^{1}{\sqrt {1-x^{2}}}dx

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

{\begin{aligned}\int {\sqrt {1-x^{2}}}dx&=\int \cos ^{2}(\theta )d\theta \\&=\int {\frac {1+\cos(2\theta )}{2}}d\theta \\&={\frac {\theta }{2}}+{\frac {\sin(2\theta )}{4}}+C={\frac {\theta }{2}}+{\frac {\sin(\theta )\cos(\theta )}{2}}+C&={\frac {\arcsin(x)+x{\sqrt {1-x^{2}}}}{2}}+C\end{aligned}}

Putting this back into the equation, we get

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

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 $b=2\pi$ .

\int \limits _{0}^{2\pi }\int \limits _{0}^{1}r\,dr\,d\theta =\int \limits _{0}^{2\pi }\left[{\frac {r^{2}}{2}}\right]_{0}^{1}d\theta =\int \limits _{0}^{2\pi }{\frac {d\theta }{2}}=\left[{\frac {\theta }{2}}\right]_{0}^{2\pi }={\frac {2\pi }{2}}=\pi

An interesting example

A less intuitive application of polar integration yields the Gaussian integral

\int \limits _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}

Try it! (Hint: multiply $\int \limits _{-\infty }^{\infty }e^{-x^{2}}dx$ and $\int \limits _{-\infty }^{\infty }e^{-y^{2}}dy$ .)

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