Area and Volume Revisited — Question 2

Question 2 – Solution

Strategy. In the first quadrant the smaller of the two polar bounds switches at the line $\theta=\pi/4$.

Step 1: Polar description

For $r\ge 0$, the disks become $$r\le 2\cos\theta,\qquad r\le 2\sin\theta.$$ Their common region lies in $0\le\theta\le\pi/2$. The bounds are equal at $\theta=\pi/4$.

Step 2: Evaluate Reflection across $y=x$ makes the two halves congruent, so $$\begin{align*} A&=2\int_0^{\pi/4}\int_0^{2\sin\theta}r\,dr\,d\theta =4\int_0^{\pi/4}\sin^2\theta\,d\theta\\ &=4\left[\frac{\theta}{2}-\frac{\sin 2\theta}{4}\right]_0^{\pi/4} =\boxed{\frac{\pi}{2}-1}. \end{align*}$$

Verification The circles have equal radius and centers $(1,0)$ and $(0,1)$, which are interchanged by reflecting across $y=x$. Their intersections, $(0,0)$ and $(1,1)$, also lie on that line.