Examples of polar integrals

Rewrite the function $f(x,y)$ in terms of $r$ and $\theta$ ,
Replace $dA$ with $r \, dr\,d\theta$ ,
Compute the limits of integration of $r$ . This means finding the minimum and maximum values of $r$ for each value of $\theta$ .
Compute the limits of integration of $\theta$ . These are numbers, not functions of $r$ , and
Do the iterated integral, one variable at a time.

In our first example, the limits of integration for $r$ do not depend on $\theta$ .

Example 1: Evaluate the integral when $D$ consists of all points $(x,\,y)$ such that

$0 \ \le \ y \ \le \sqrt{9-x^2}\,, \quad 0\ \le \ x \ \le 3\,.$

We worked this example in the last section using rectangular coordinates. It is substantially easier in polar coordinates. Our region is the first quadrant inside a circle of radius 3, as shown to the right. This means that our limits of integration are that $r$ goes from 0 to 3 and $\theta$ goes from $0$ to $\pi/2$ . Our function is

$f(x,y)\ =\ x+y \ = \ r\cos(\theta) + r\sin(\theta).$ So the double integral

$I$ becomes the iterated integral

$\int_0^{\pi/2}\! \int_0^3 r (\cos(\theta) + \sin(\theta)) r\, dr\, d\theta.$ For fixed

$\theta$ , the integral over

$r$ is easy:

$\int_0^3 r^2 [\cos(\theta) + \sin(\theta)] dr = 9 [\cos(\theta) + \sin(\theta)].$

Thus

$I\ = \ \int_0^{\pi/2}\, 9(\cos(\theta)+\sin(\theta)) d\theta \ = \ 18\,.$ Unlike the integral in rectangular coordinate, there are no square roots and no fancy trig substitutions are required. We just have a simple integral.

When our regions are more complicated than circles, we have to be more careful.

Example 2: Evaluate the integral when $D$ consists of all points above the $x$ axis and inside the cardioid $r = 1+\cos(\theta)$ .

Our function is $f(x,y) = y = r\sin(\theta)$ . For fixed $\theta$ , $r$ goes from $0$ to $1 + \cos(\theta)$ . Since we are above the $x$ axis, $\theta$ runs from 0 to $\pi$ . Thus our double integral becomes the iterated integral

$\int_0^{\pi}\! \int_0^{1+\cos(\theta)} r \sin(\theta) r\, dr\, d\theta.$ For fixed

$\theta$ , the integral over

$r$ is

$\int_0^{1+\cos(\theta)} r^2 \sin(\theta) dr = \frac{\sin(\theta)(1+\cos(\theta))^3}{3}.$

Thus

$\begin{eqnarray*} I & = & \int_0^{\pi}\, \frac{\sin(\theta)(1+\cos(\theta))^3}{3} \, d\theta \\ &=& \left . - \frac{(1+\cos(\theta))^4}{12} \right |_0^\pi \ = \ \frac{4}{3}. \end{eqnarray*}$

The function $e^{-x^2}$ does not have an antiderivative that can be expressed in closed form, and it is impossible to compute $\int_{0}^1 e^{-x^2} dx$ exactly. However, the integral $\displaystyle{\int_{-\infty}^\infty e^{-x^2} dx}$ turns out to equal $\sqrt{\pi}.$ In the following video, we use double integrals and polar coordinates to explain this surprising result.

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Modeling with Differential Equations

Euler's Method and Direction Fields

Separable Equations

Models of Growth

Linear Equations

Parametrized Curves

Calculus with Parametrized Curves

Polar Coordinates

Areas and Lengths of Polar Curves

Conic sections

Conic Sections in Polar Coordinates

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Functions of 2 and 3 variables

Partial Derivatives

Differentiability and the Chain Rule

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Double integrals in polar coordinates

Multiple integrals in physics

Integrals in Probability and Statistics

Change of Variables