15. Polar Coordinates

d. Area Enclosed by a Polar Graph

2. Area Between Polar Graphs

So far, we have seen how to compute the area between a polar curve and the origin. We now look at computing the area between two polar curves.

Find the area bounded by the graphs of \(r=f(\theta)\) and \(r=g(\theta)\) between the rays \(\theta=\alpha\) and \(\theta=\beta\). Assume \(f(\theta) \ge g(\theta) \ge 0\).

This plot shows two polar curves on the coordinate axes. The outer curve
is labeled r = f of theta and the inner curve is labeled r = g of theta.
Both look like pieces of ellipses.
There are two rays from the origin labeled theta = alpha and theta = beta,
forming an area between the two curves and between the two rays.

The area enclosed by \(r=f(\theta)\) is \(\displaystyle A=\int_\alpha^\beta \dfrac{1}{2}f(\theta)^2\,d\theta\), while the area enclosed by \(r=g(\theta)\) is \(\displaystyle A=\int_\alpha^\beta \dfrac{1}{2}g(\theta)^2\,d\theta\). We simply subtract to obtain:

The area bounded by the graphs of \(r=f(\theta)\) and \(r=g(x)\) between the rays \(\theta=\alpha\) and \(\theta=\beta\) is \[ A=\int_\alpha^\beta \dfrac{1}{2}[f(\theta)^2-g(\theta)^2]\,d\theta \] provided \(f(\theta) \ge g(\theta) \ge 0\).

Find the area of the region outside the circle \(r=1\) but inside the cardioid, \(r=2+2\sin\theta\).

This plot shows two polar curves.
One curve is the circle centered at the origin with radius 1. The other
curve is the cardioid r = 2 plus 2 sine theta, a heart-shaped curve that
lies mostly above the x axis and has a cusp at the origin at the bottom
of the heart. They intersect in the third and fourth quadrants.

The only difficulty is finding the limits of integration. These are where the two curves intersect. So we equate the functions and solve: \[ 2+2\sin\theta=1 \qquad \Rightarrow \qquad 2\sin\theta=-1 \qquad \Rightarrow \qquad \sin\theta=-\,\dfrac{1}{2} \] The intersection in the \(4^\text{th}\) quadrant is at \(\theta=-\,\dfrac{\pi}{6}\). The intersection in the \(3^\text{rd}\) quadrant is at \(\theta=\dfrac{7\pi}{6}\). So the area is \[\begin{aligned} A &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [(2+2\sin\theta)^2-1^2]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [3+8\sin\theta+4\sin^2]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [3+8\sin\theta+2(1-\cos(2\theta))]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [5+8\sin\theta-2\cos(2\theta)]\,d\theta \\ &=\dfrac{1}{2}[5\theta-8\cos\theta-\sin(2\theta)]_{-\pi/6}^{7\pi/6} =\dfrac{10\pi}{3}+\dfrac{7\sqrt{3}}{2} \end{aligned}\]

It is important to integrate from \(\theta=-\,\dfrac{\pi}{6}\) to \(\theta=\dfrac{7\pi}{6}\) which covers the area between the curves, including \(\theta=0\), \(\dfrac{\pi}{2}\) and \(\pi\). If one instead integrates from \(\theta=\dfrac{7\pi}{6}\) to \(\theta=\dfrac{11\pi}{6}\), then one gets the area at the bottom inside the circle outside the cardioid.

Caution: It is not always possible to find the points of intersection of two polar graphs by simply equating the functions as was done in the preceeding example. Recall the discussion on Intersections of Polar Graphs.

Find the area of the region enclosed by both polar graphs, \(r=2\sin\theta\) and \(r=2\cos\theta\).

The graphs are circles. By symmetry, you can compute half of the area and double it.

This plot shows two circles with radius 1. One circle is centered at
x = 0 and y = 1, and the other circle is centered at x = 1 and y = 0.
The circles overlap in the first quadrant, and a dashed line connects the origin
to the intersection point at x = 1 and y = 1.

\(A=\dfrac{\pi}{2}-1\)

The graphs are circles. In a previous exercise, we found the graphs intersect at \((r,\theta)=\left(\sqrt{2},\dfrac{\pi}{4}\right)\) and at the origin. By symmetry, we split the region in half (as shown) and double the area. The bottom half is bounded by the ray at \(\theta=\dfrac{\pi}{4}\) and the red circle \(r=2\sin\theta\) which comes into the origin at \(\theta=0\). So the area is:

\[\begin{aligned} A&=2\int_0^{\pi/4} \dfrac{1}{2}r(\theta)^2\,d\theta =2\int_0^{\pi/4} \dfrac{1}{2}(2\sin\theta)^2\,d\theta \\ &=4\int_0^{\pi/4} \sin^2\theta\,d\theta =2\int_0^{\pi/4} (1-\cos(2\theta))\,d\theta \\ &=2\left[\theta-\,\dfrac{\sin(2\theta)}{2}\right]_0^{\pi/4} =\dfrac{\pi}{2}-1 \end{aligned}\]

15. Polar Coordinates

d. Area Enclosed by a Polar Graph

2. Area Between Polar Graphs

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