Evaluate the double integral of x + 2y over the region where 1 ≤ x² + y² ≤ 4, y ≥ x, and x ≤ 0.

Step-by-step solution

1. 1

   Switch to polar because the region is bounded by circles. Set

   $x = r\cos\theta, \qquad y = r\sin\theta, \qquad dA = r\,dr\,d\theta$

   The annulus condition becomes simply $1 \le r \le 2$, independent of $\theta$ — that separation is what makes polar worth the change of variables here.

2. 2

   Translate the first angular condition. $x \le 0$ means $r\cos\theta \le 0$, and since $r > 0$:

   $\cos\theta \le 0 \ \Longrightarrow \ \theta \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$

   This is the left half-plane.

3. 3

   Translate the second angular condition. $y \ge x$ means $\sin\theta \ge \cos\theta$. Writing the difference as a single sine,

   $\sin\theta - \cos\theta = \sqrt{2}\,\sin\!\left(\theta - \frac{\pi}{4}\right) \ge 0 \ \Longrightarrow \ \theta \in \left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$

   Intersecting the two ranges gives the wedge

   $\frac{\pi}{2} \le \theta \le \frac{5\pi}{4}$

   an angular span of $3\pi/4$. Getting this intersection wrong is the main hazard in the problem.

4. 4

   Convert the integrand and pick up the Jacobian.

   $(x + 2y)\,dA = r(\cos\theta + 2\sin\theta)\cdot r\,dr\,d\theta = r^2(\cos\theta + 2\sin\theta)\,dr\,d\theta$

   The extra factor of $r$ from $dA$ is what turns $r^1$ into $r^2$ — omitting it is the classic polar mistake.

5. 5

   The integral separates; do the radial part. Since the $r$-limits do not depend on $\theta$:

   $\int_1^2 r^2\,dr = \left[\frac{r^3}{3}\right]_1^2 = \frac{8 - 1}{3} = \frac{7}{3}$

6. 6

   Do the angular part and combine.

   $\int_{\pi/2}^{5\pi/4} (\cos\theta + 2\sin\theta)\,d\theta = \bigl[\sin\theta - 2\cos\theta\bigr]_{\pi/2}^{5\pi/4}$

   At $\theta = 5\pi/4$ both sine and cosine equal $-\frac{\sqrt2}{2}$, giving $-\frac{\sqrt2}{2} + \sqrt2 = \frac{\sqrt2}{2}$; at $\theta = \pi/2$ the bracket is $1 - 0 = 1$. So the angular integral is $\frac{\sqrt2}{2} - 1$, and

   $\iint_D (x+2y)\,dA = \frac{7}{3}\left(\frac{\sqrt{2}}{2} - 1\right) = \frac{7}{6}\left(\sqrt{2} - 2\right) \approx -0.6834$

   A negative value is expected: the wedge sits mostly where $x < 0$. Quadrature in polar gives $-0.68341751$, and a Monte-Carlo check on the raw Cartesian conditions gives $-0.681$.