Double Integrals in Polar Coordinates — Question 8

Question 8 – Solution

Strategy. The desired petal begins and ends where $\cos(2\theta)=0$ around $\theta=0$.

Step 1: Bounds The adjacent zeros are $\theta=\pm\pi/4$, and $\cos(2\theta)\ge 0$ between them. Thus $0\le r\le\cos(2\theta)$.

Step 2: Area $$\begin{align*} A&=\int_{-\pi/4}^{\pi/4}\int_0^{\cos 2\theta}r\,dr\,d\theta =\frac 12\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)d\theta\\ &=\frac 14\left[\theta+\frac{\sin 4\theta}{4}\right]_{-\pi/4}^{\pi/4} =\boxed{\frac\pi 8}. \end{align*}$$

Verification The sine endpoint terms vanish. Extending beyond these zeros introduces negative radial values, which trace different petals rather than enlarging the positive-$x$ petal.