Double Integrals in Polar Coordinates — Question 4

Question 4 – Solution

Strategy. Solve $2\cos\theta=1$ to locate the two intersection rays, then integrate from the unit circle to the offset circle.

Step 1: Angles The curves meet when $\cos\theta=1/2$, so $\theta=\pm\pi/3$. Between these angles, $2\cos\theta\ge 1$.

Step 2: Area $$\begin{align*} A&=\int_{-\pi/3}^{\pi/3}\int_1^{2\cos\theta}r\,dr\,d\theta\\ &=\frac 12\int_{-\pi/3}^{\pi/3}(4\cos^2\theta-1)d\theta\\ &=\frac 12\left(4\left(\frac\pi 3+\frac{\sqrt 3}{4}\right)-\frac{2\pi}{3}\right) =\boxed{\frac\pi 3+\frac{\sqrt 3}{2}}. \end{align*}$$

Verification At both endpoint rays the inner and outer radii equal $1$, so slice width vanishes. At $\theta=0$ the slice runs from $1$ to $2$, confirming the order.