Question

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

Short answer

The area of the common region is \(a^2 [\frac{\pi}{2} - 1]\)

Step-by-step solution

Find the Points of Intersection

Set the two polar equations equal to each other and solve for \( \theta \). So, solve \( a(1+cos \theta) = a sin \theta \). This yields \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{7\pi}{4} \)

Plot the Polar Coordinates

Plot the functions to see where their regions overlap, which areas need to be integrated. Here, it can be observed that their mutual region falls within the range \( \theta = \frac{\pi}{4} \) to \( \theta = \frac{7\pi}{4} \)

Use Double Integration to Find the Area

Determine which curve is the outer boundary (\(a(1+cos \theta)\)) and which is the inner boundary (\(a sin \theta\)), and form the integral for the difference between the two from \( \frac{\pi}{4} \) to \( \frac{7\pi}{4} \). So, the integral formed will be \(\int^{7\pi/4}_{\pi/4} \int^{a(1+cos\theta)}_{a sin \theta} r dr d\theta \)

Calculate the Value of the Integral

By changing bounds, compute the integral to attain the final numerical answer to the area. Therefore, the area becomes \(\int^{7\pi/4}_{\pi/4} \frac{1}{2} [r^2]^{a(1+cos\theta)}_{a sin \theta}d\theta \). On computing, the result is \(Area = a^2 [\frac{\pi}{2} - 1]\)