Question

Find the area of the region. Common interior of \(r=a \cos \theta\) and \(r=a \sin \theta\) where \(a>0\)

Short answer

The area of the region is \( a^2 \).

Step-by-step solution

Find the Intersection Points

To find the intersection points, set \( a \cos \theta \) equal to \( a \sin \theta \), i.e., \( a \cos \theta = a \sin \theta \), which simplifies to \( \tan \theta = 1 \). Hence, the solutions are \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{5\pi}{4} \). These will be the limits of the integration.

Set Up the Integral

We want to calculate the area of common interior region of \( \theta = r a \cos \theta \) and \( \theta = r a \sin \theta \). So The area \( A \) is given by the integral \( A = \frac{1}{2} \int_{\pi/4}^{5\pi/4} [(a \cos \theta)^2 - (a \sin \theta)^2] d \theta \) .

Solve the Integral

Evaluating the integral, we get \( A = \frac{1}{2} \int_{\pi/4}^{5\pi/4} a^2 [\cos^2 \theta - \sin^2 \theta] d \theta \). Using the double angle formula \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \), this simplifies to \( A = \frac{1}{2} a^2 \int_{\pi/4}^{5\pi/4} \cos 2\theta d \theta \). Evaluating this integral gives \( A = a^2 \).