Double Integrals in Polar Coordinates — Question 7

Question 7 – Solution

Strategy. Rewrite the circle as $r=2\cos\theta$ and remember that $x=r\cos\theta$ contributes another factor of $r$ with $dA$.

Step 1: Bounds The boundary equation becomes $r^2-2r\cos\theta=0$. Thus $$-\frac\pi 2\le\theta\le\frac\pi 2,\qquad 0\le r\le 2\cos\theta.$$

Step 2: Evaluate $$\begin{align*} \iint_Dx\,dA &=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}r^2\cos\theta\,dr\,d\theta\\ &=\frac 83\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta =\frac 83\cdot\frac{3\pi}{8}=\boxed{\pi}. \end{align*}$$

Verification The disk has area $\pi$ and is symmetric about the vertical line $x=1$, so its average $x$-coordinate is $1$. Hence its first moment is $1\cdot\pi=\pi$.