Question

Find the area of the region. Inside \(r=a(1+\cos \theta)\) and outside \(r=a \cos \theta\)

Short answer

The area of the region inside the curve \(r=a(1+\cos \theta)\) and outside the curve \(r=a \cos \theta\) is \(\frac{a^2\pi}{2}\)

Step-by-step solution

Find the Points of Intersection

Set the two equations equal to each other to find the common points: \(a(1 + \cos \theta) = a \cos \theta\). This simplifies to \(1 + \cos \theta = \cos \theta\), which gives \(\cos \theta = 1\) and the corresponding \(\theta\) values are \(0\) and \(\pi\).

Set Up the Integral

The area A of a region in polar coordinates is given by \(\frac{1}{2}\int \left( r_2^2 - r_1^2 \right) d\theta\), where \(r_2\) is the outside curve and \(r_1\) is the inside curve, and integrate from the smaller \(\theta\) value to the larger \(\theta\) value. Here, \(r_2 = a(1+\cos \theta)\) and \(r_1 = a cos \theta\), and \(\theta\) ranges from 0 to \(\pi\), therefore, the formula for A becomes:\(A =\frac{1}{2}\int_0^\pi \left[ a^2(1+\cos \theta)^2 - a^2 \cos^2 \theta \right] d\theta\).

Evaluate the Integral

Using the integral rules and properties, this integral simplifies to: \(A =\frac{1}{2}a^2 \int_0^\pi{(1 + 2\cos \theta +\cos^2 \theta)- \cos^2 \theta}d\theta = \frac{1}{2}a^2\int_0^\pi{1 + 2\cos \theta}d\theta = \frac{1}{2}a^2(1+2(\sin \pi - \sin 0)) =\frac{a^2\pi}{2}\)