Question

Find the area of the region. One petal of \(r=2 \cos 3 \theta\)

Short answer

\[A = \pi/3 - 1/2 \]

Step-by-step solution

Find limits of integration

To find the limits of integration that define a single petal, set the equation of the polar curve, \(r=2 \cos 3 \theta = 0\). This yields \(\cos 3 \theta = 0\), so \(3 \theta = \frac{\pi}{2}\) or \(3 \theta = \frac{3 \pi}{2}\). Dividing both results by 3 finds the limits of \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{\pi}{2}\).

Apply Formula for Area

The formula for area in polar coordinates is \[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d \theta\]. Substituting in the equation for \(r\) and the limits of \(\theta\) finds that \[A = \frac{1}{2} \int_{\pi/6}^{\pi/2} (2 \cos 3 \theta)^2 d \theta\]

Integrate the Equation

First simplify the integrand: \[A = \frac{1}{2} \int_{\pi/6}^{\pi/2} 4 \cos^2 3 \theta d \theta = 2 \int_{\pi/6}^{\pi/2} \cos^2 3 \theta d \theta\]. Then use the power-reduction identity \(\cos^2 a = \frac{1+ \cos 2a}{2}\) to rewrite the integral: \[A = 2 \int_{\pi/6}^{\pi/2} \frac{1+ \cos 6 \theta}{2} d \theta = \int_{\pi/6}^{\pi/2} (1+ \cos 6 \theta) d \theta\]

Compute the Definite Integral

Integrate to get \[A = \theta+ \frac{1}{6} \sin 6 \theta | _{\pi/6}^{\pi/2}\] The integral evaluated at the upper limit \(\pi/2\) is \( \pi/2 + 0 \), and evaluated at the lower limit \(\pi/6\) is \(\pi/6 + 1/2\). Taking the difference, the final result is \[A = ( \pi/2 + 0) - ( \pi/6 + 1/2) = \pi/3 - 1/2\]