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Chapter 9: Problem 29

Find the area of the region. Common interior of \(r=a(1+\cos \theta)\) and \(r=a \sin \theta\)

Short Answer

Expert verified
The area of the common region is \(\frac{a^2}{8}\pi\).

Step by step solution

01

Plot the Polar Curves

To start off, plot the polar curves \(r = a(1+\cos(\theta))\) and \(r = a\sin(\theta)\). The first curve is a cardioid while the second curve is a circle of radius a centered at the origin.
02

Find the Intersection Points

In order to evaluate the integral from the correct bounds, we have to determine the intersection points of the two curves. Solve the system of equations \(a(1+\cos \theta) = a \sin \theta\). From this equation, we solve for two values of theta, \(0\) and \(\frac{\pi}{2}\).
03

Setup the Integral for Area Calculation

Once we have the polar coordinates of the intersection points, we can use them to set up the integral. The formula for area in polar coordinates is \(\frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\). We observe that within the given bounds, let's say from \(\alpha\) to \(\beta\), we have \(r = a\sin(\theta)\) inside \(r = a(1+\cos\theta)\).
04

Evaluate the Integral

Now, let's calculate the area. Substitute \(r = a\sin(\theta)\) and evaluate the integral: \(Area = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (a\sin(\theta))^2 d\theta = \frac{a^2}{2}\int_{0}^{\frac{\pi}{2}} \sin^2(\theta) d\theta \). Using the rule that \(\int \sin^2(\theta) d\theta = \frac{\theta}{2} - \frac{\sin(2\theta)}{4}\), we finally get: \(Area = \frac{a^2}{2}[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}]_{0}^{\frac{\pi}{2}} = \frac{a^2}{8}\pi\).

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