When using rectangular coordinates, the equations \(x=h\) and \(y=k\) defined vertical and horizontal lines, respectively, and combinations of these lines create rectangles (hence the name “rectangular coordinates”). It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral.

When using polar coordinates, the equations \(\theta =\alpha \) and \(r=c\) form lines through the origin and circles centered at the origin, respectively, and combinations of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles. Recall that the area of a sector of a circle with radius \(r\) subtended by an angle \(\theta \)

is \(A = \frac {1}{2}\theta r^2\). So given a polar plot, partition the interval \([\alpha ,\beta ]\) into \(n\) equally spaced subintervals as \(\alpha = \theta _1 < \theta _2 <\cdots <\theta _{n+1}=\beta \):

The length of each subinterval is \(\Delta \theta = (\beta -\alpha )/n\), representing a small change in angle. The area of the region defined by the \(i\)th subinterval \([\theta _i,\theta _{i+1}]\) can be approximated with a sector of a circle with radius \(r(\theta ^*_i)\), for some \(\theta ^*_i\) in \([\theta _i,\theta _{i+1}]\). The area of this sector is \(\frac 12r(\theta ^*_i)^2\Delta \theta \). This is shown here

where \([\alpha ,\beta ]\) has been divided into \(4\) subintervals. We approximate the area of the whole region by summing the areas of all sectors:

This is a Riemann sum! By taking the limit of the sum as \(n\to \infty \), we find the exact area of the region in the form of a definite integral.

The theorem states that \(0\leq \beta -\alpha \leq 2\pi \). This ensures that region does not overlap itself, giving a result that does not correspond directly to the area.

Compute the area of one petal of the polar curve \(r(\theta ) = \cos (3\theta )\):

From the picture it looks like integrating from \(\theta =-\pi /6\) to \(\pi /6\) will give us the area of our desired region. We can convince ourselves that this is correct by inspecting

\[ r(\theta ) = \cos (3\theta ) \]

and noting that our curve starts at \((\answer [given]{1},\answer [given]{0})\) in the \((x,y)\)-plane when \(\theta =0\), and then moves to the origin. Since the first positive value of \(\theta \) that make \(r(\theta )= 0\) is \(\answer [given]{\pi /6}\), we see that our eyes did not lie to us. Write with me \begin{align*} \frac {1}{2}\int _{-\pi /6}^{\pi /6} r(\theta )^2 d\theta &= \frac {1}{2}\int _{-\pi /6}^{\pi /6} \cos ^2(3\theta ) d\theta \\ &= \frac {1}{2}\int _{-\pi /6}^{\pi /6} \frac {1-\cos (6\theta )}{2} d\theta \\ &= \frac {1}{4}\int _{-\pi /6}^{\pi /6} 1-\cos (6\theta ) d\theta \\ &= \frac {1}{4}\bigg [\theta -\frac {\sin (6\theta )}{6}\bigg ]_{-\pi /6}^{\pi /6}\\ &= \answer [given]{\frac {\pi }{12}}. \end{align*}

Now I present you with a mystery: You know that the area of one petal of \(r(\theta ) = \cos (3\theta )\) has area \(\pi /12\). Hence, all three petals must have an area of \(\pi /4\). Compare this to \begin{align*} \frac {1}{2}\int _0^{2\pi } \cos ^2(3\theta ) d\theta &= \frac {1}{2}\int _{0}^{2\pi } \frac {1-\cos (6\theta )}{2} d\theta \\ &= \frac {1}{4}\int _{0}^{2\pi } 1-\cos (6\theta ) d\theta \\ &= \frac {1}{4}\int _{0}^{2\pi } \theta -\frac {\sin (6\theta )}{6} d\theta \\ &= \frac {1}{4}\bigg [\theta -\frac {\sin (6\theta )}{6}\bigg ]_{0}^{2\pi }\\ &= \frac {\pi }{2} \end{align*}

How is it possible that the total area of the three petals is \(\pi /4\), but the integral above is \(\pi /2\)?

one (or more!) of our computations has a mistake \(\pi /4 = \pi /2\) a complete curve is drawn when \(\theta \) runs from \(0\) to \(\pi \)

A good way to understand the polar graph is to graph \(y=r(x)\) along side of \(r(\theta )\):

1 Areas between polar curves

Consider the shaded region:

If we let \(r(\theta )\) represent the circle, and \(s(\theta )\) represent the cardioid, we can find the area of this region by computing the area bounded by \(r(\theta )\) and subtracting the area bounded by \(s(\theta )\) on \([\alpha ,\beta ]\). Thus \begin{align*} \text {Area} &= \frac 12\int _\alpha ^\beta r(\theta )^2 d\theta -\frac 12\int _\alpha ^\beta s(\theta )^2 d\theta \\ &= \frac 12\int _\alpha ^\beta \big (r(\theta )^2-s(\theta )^2\big ) d\theta . \end{align*}

Find the area of the region that lies within

\[ r(\theta ) = 1 \]

but not within

\[ s(\theta ) = 1-\sin (\theta ) \]

This corresponds to the following region:

Hence to find the are we compute \begin{align*} \frac {1}{2}\int _0^{\pi } r(\theta )^2 - s(\theta )^2 d\theta &= \frac {1}{2}\int _0^{\pi } 1 - \left (1-\sin (\theta )\right )^2 d\theta \\ &= \frac {1}{2}\int _0^{\pi } 1 - \left (1 - 2\sin (\theta ) + \sin ^2(\theta )\right ) d\theta \\ &= \frac {1}{2}\int _0^{\pi } 1 -1 + 2\sin (\theta ) - \sin ^2(\theta ) d\theta \\ &= \frac {1}{2}\int _0^{\pi } 2\sin (\theta ) - \sin ^2(\theta ) d\theta \\ &= \frac {1}{2}\int _0^{\pi } 2\sin (\theta ) - \frac {1-\cos (2\theta )}{2} d\theta \\ &= \frac {1}{4}\int _0^{\pi } 4\sin (\theta ) - 1+\cos (2\theta ) d\theta \\ &= \frac {1}{4}\bigg [-4\cos (\theta ) - \theta +\frac {\sin (2\theta )}{2}\bigg ]_0^\pi \\ &= \answer [given]{2-\frac {\pi }{4}}. \end{align*}

When computing regions between polar curves, one must be careful. Sometimes, one must compute the points of intersection of the two curves, and this can be difficult.

Find the points of intersection between the curves \begin{align*} r(\theta ) &= 1+ \cos (\theta )\\ s(\theta ) &= 1+ \sin (\theta ). \end{align*}

To find the points of intersection of two polar curves, the best thing to do is to look at a graph.

From the graph above, we see that there are \(3\) points of intersection. Let’s try to use algebra to find them. Write with me \begin{align*} 1+ \cos (\theta ) &= 1+ \sin (\theta )\\ \cos (\theta ) &=\sin (\theta ) \end{align*}

Ah, and this is true when:

\[ \theta = \answer [given]{\pi /4} \qquad \text {and}\qquad \theta = \answer [given]{5\pi /4} \]

However, from the graph we see there is one more point of intersection. Namely

\[ r(\pi ) = s\left (\answer [given]{3\pi /2}\right ) = 0. \]

When you are unable to look at a graph, you can still use algebra to try and find the points of intersection, but some may be hidden.

Find the area of the region that lies within both polar curves \begin{align*} r(\theta ) &= 1+ \cos (\theta )\\ s(\theta ) &= 1+ \sin (\theta ). \end{align*}

To compute this area, we’ll use the previous example. First we’ll compute the area of this region:

So to compute the region we can either compute:

\[ \int _{\pi /4}^{5\pi /4} (1+\cos (\theta ))^2 d\theta \qquad \text {or}\qquad \int _{-3\pi /4}^{\pi /4} (1+\sin (\theta ))^2 d\theta \]

Evaluating either integral tells us the area is \(\answer [given]{3\pi -4\sqrt {2}}\).