Double Integrals in Polar Coordinates
Polar coordinates and polar curves
Besides the usual Cartesian coordinates , we can also specify a point in 2D space using polar coordinates , where is the signed distance from the origin , usually called the pole, to the point and is the angle measured (anti-clockwise if and clockwise if ) from the positive x-axis to the ray from the pole toward if or to the opposite ray if .
In the applet below, you can input various values of and to find out the position of the point . Moreover, you can verify that the polar coordinates and represent the same point.
If you are given the polar coordinates , you can convert them to the usual Cartesian coordinates using the following well-known formulas:
On the other hand, if you are given the Cartesian cooridnates , you can find the polar coordinates using the following equations:
Polar equations and polar curves
A polar equation is an equation in and . The set of points in 2D space whose polar coordinates satisfy such equation is the polar curve of the equation. In fact, for some curves, especially the ones that are closed loops, they can be more conveniently described by polar equations.
Basic examples:
Circle centered at the origin with radius :
Its polar equation is .
A line through the origin such that the angle between the line and the x-axis is :
Its polar equation is .
The circle , where :
We can obtain its polar equation by substitutions and as follows:
(Note: Similarly, the circle has the polar equation .)
More interesting examples:
Archimedean spiral
Polar equation: , where
Cardioid (a heart-shaped curve)
Polar equation:
Rose family
Polar equation: , where are positive integers
In the applet below, you can plot the above polar curves and trace the curves using the slider.
Double integrals in polar coordinates
Given a function of two variables defined on a region , we already learned the definition of its double integral over , i.e. . If is the kind of regions like sectors or rotationally symmetric shapes, which can be more easily described using polar coordinates, the double integral would be much easier to compute if we can rewrite it in terms of polar coordinates.
Suppose we consider , the polar rectangular region, as shown in the applet below. We divide into subintervals of equal length . Also, we divide into subintervals of equal length . The arcs and rays divide into polar rectangles.
Let be the centre of the polar rectangle, then the volume under the the graph of can be approximated by
where and is the area of the polar rectangle.
To compute , we use the formula for the area of the sector - , where and are the radius and the angle (in radian) of the sector respectively. Therefore, we have
(Note: Conceptually, can be regarded as .)
Taking , we have
And by Fubini's theorem, we have
In the applet below, you can change the range of and to obtain a different polar rectangular region. Observe that the small polar rectangle is getting larger when it is further away from the origin. The factor "" in the corresponds to such scaling.
Example:
Find the volume of a sphere with radius using double integral.
Answer:
We may assume that the sphere is centered at the origin. Then the equation of the sphere is
By symmetry, it suffices to consider the volume enclosed by the upper hemisphere and the xy-plane. Then the upper hemisphere is the graph of the function on the circular disk , i.e. .
Using polar coordinates, we have and the volume of the sphere can be expressed as the following double integral:
Exercise: Find the volume beneath the graph of and above the annular region .
Double integrals over general polar regions
Given two polar curves with equations and such that for . The polar region is defined as follows:
In other words, is the region bounded by the polar curve of as the inner boundary and as the outer boundary, where the angle sweeps from to . Then we can express the double integral of the function over as the following iterated integral:
Example: Let on , the region in the first quadrant bounded by the circle and the cardioid . Find the value of .
Answer:
Using polar coordinates, we have .
Since for (in the first quadrant), the circle is the inner boundary and the cardioid is the outer boundary.
Therefore, .
The double integral in polar coordinates is as follows:
In the applet below, you can see the shape of the region and the solid under the graph over the region, whose volume is the value of the double integral. In the left panel, the red line segment is the "slice" of the region corresponding to a fixed value of . The cross section of the solid under the graph along the red line segment corresponds to the integral .
Exercise: Let be a region such that it is outside and inside . Suppose on . Express the volume of the solid under the graph of over as an iterated integral. (You can use the above applet to find out how the region looks like.)
An application
The following is the Gaussian integral, which is a very important integral in statistics:
There is no elementary way to calculate the indefinite integral . But the above improper integral can be evaluated using the famous "polar coordinates trick".
First of all, is an even function, we have
i.e. It suffices to compute .
Let and , i.e., the first quadrant. Consider the following double integral:
We can compute it as follows:
Observe that the above double integral can also be computed using Cartesian coordinates:
Therefore, we have
Hence, the Gaussian integral .