Question

To find the area of the region that lies inside both curves.

Short answer

The area of the region that lies inside both the curves is \(\frac{{3\pi }}{2} - 4\).

Step-by-step solution

Given data

The given polar equations are shown below.

\(r = 1 + \cos \theta \) …… (1)

\(r = 1 - \cos \theta \) …… (2)

Concept of Area of Polar region

The area\(A\)of the polar region\(R\)is found by\(A = \frac{1}{2}\int_a^b {{r^2}} d{\theta ^n}\).

Sketch the graph of the curves

Use online graphing calculator and sketch the graph of \(r = 1 + \cos \theta \) and \(r = 1 - \cos \theta \) as shown in Figure below.

From figure, it is observed the shaded region is symmetric about the \(y\) axis.

Note that \( - \frac{\pi }{2}\) and \(\frac{\pi }{2}\) are the angles where the two curves intersect.

Therefore, the limit of integration is from \( - \frac{\pi }{2}\) to \(\frac{\pi }{2}\).

Calculation of the area of the region

Calculate the area of the region by the polar area formula.

\(A = 2\int_a^b {\frac{{{r^2}}}{2}} d\theta \) …… (3)

Substitute \(1 - \cos \theta \) for \(r\) in the equation (3).

\(\begin{aligned}{l}A = 2\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{{(1 - \cos \theta )}^2}}}{2}} d\theta \\A = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + {{\cos }^2}\theta - 2\cos \theta } \right)} d\theta \end{aligned}\)

Substitute \(\frac{{1 + \cos 2\theta }}{2}\) for \({\cos ^2}\theta \).

\(\begin{aligned}{l}A = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + \frac{{1 + \cos 2\theta }}{2} - 2\cos \theta } \right)} d\theta \\A = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {\frac{{2 + 1 + \cos 2\theta - 4\cos \theta }}{2}} \right)} d\theta \\A = \frac{1}{2}\int_{ - \frac{\pi }{2}}^{\frac{x}{2}} {(3 + \cos 2\theta - 4\cos \theta )} d\theta \end{aligned}\)

Integrate the above equation of the area of the region of curve

Integrate the above equation with respect to \(\theta \).

\(A = \frac{1}{2}\left( {3\theta + \frac{1}{2}\sin 2\theta - \left. {4\sin \theta } \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}} \right.\)

Apply the limits.\(\begin{aligned}{l}A = \frac{1}{2}\left( {\left( {3\left( {\frac{\pi }{2}} \right) + \frac{1}{2}\sin 2\left( {\frac{\pi }{2}} \right) - 4\sin \left( {\frac{\pi }{2}} \right)} \right)} \right.\left. { - \left( {3\left( {\frac{{ - \pi }}{2}} \right) + \frac{1}{2}\sin 2\left( {\frac{{ - \pi }}{2}} \right) - 4\sin \left( {\frac{{ - \pi }}{2}} \right)} \right)} \right)\\A = \frac{1}{2}\left( {\left( {\frac{{3\pi }}{2} - 4} \right) - \left( {\frac{{ - 3\pi }}{2} + 4} \right)} \right)\\A = \frac{1}{2}\left( {\frac{{3\pi }}{2} - 4 + \frac{{3\pi }}{2} - 4} \right)\\A = \frac{1}{2}(3\pi - 8)\end{aligned}\)

Therefore, the area of the region that lies inside both the curves is \(\frac{{3\pi }}{2} - 4\).