Question

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

Short answer

The area of one petal of the polar curve \(r=\cos 2 \theta\) is \(\frac{\pi}{16}\).

Step-by-step solution

Identify the range over which the petal exists

Observe the curve and identify where one petal of the curve occurs. For \(r = \cos 2\theta\), a full petal occurs where the angle \(\theta\) changes from 0 to \(\pi/4\). Therefore, \(\alpha = 0\) and \(\beta = \pi/4\).

Insert into the area formula

Now that we have our limits \(\alpha\) and \(\beta\), we can insert them into the polar area formula along with our function for \(r\). This results in: \(\text{Area} = \frac{1}{2} \int_0^{\pi/4} (\cos2\theta)^2 \,d\theta\).

Perform the integration

Perform the integration which simplifies to: \(\text{Area} = \frac{1}{2} \int_0^{\pi/4} \frac{1+\cos4\theta}{2} \,d\theta\). This integral can be split into two simpler integrals: \(\text{Area} = \frac{1}{2} \int_0^{\pi/4} \frac{1}{2} d\theta + \frac{1}{2} \int_0^{\pi/4} \frac{\cos4\theta}{2} d\theta\). Now, integrate each individual integral: \(\text{Area} = \frac{1}{2} [\frac{\theta}{2}]_0^{\pi/4} + \frac{1}{2} [\frac{\sin 4\theta}{8}]_0^{\pi/4}\).

Evaluate the definite integrals

Evaluating the integrals at their respective bounds sources: \(\text{Area} = \frac{1}{2} [(\frac{\pi}{4*2}) - (\frac{0}{2})] + \frac{1}{4*2} [(\sin \pi) - (\sin 0)] = \frac{\pi}{16}\). Note that the second portion of the area calculation equals zero, as the sine functions both evaluate to zero.