Double Integrals in Polar Coordinates — Question 2

Question 2 – Solution

Strategy. The factor $1/r$ cancels the polar Jacobian, leaving a rectangular integral in $(r,\theta)$.

Step 1: Bounds The radial inequality gives $1\le r\le 3$, while $\pi/6\le\theta\le\pi/2$.

Step 2: Evaluate $$\begin{align*} \iint_D\frac 1{\sqrt{x^2+y^2}}dA &=\int_{\pi/6}^{\pi/2}\int_1^3\frac 1r(r\,dr\,d\theta)\\ &=\int_{\pi/6}^{\pi/2}2\,d\theta=\boxed{\frac{2\pi}{3}}. \end{align*}$$

Verification The region stays at least one unit from the origin, so the integrand is bounded and no singularity occurs. After cancellation the radial length is $2$ and the angular width is $\pi/3$.