The value of the double integral $I = \iint_D \sqrt{x^2 + y^2} \, dx \, dy$, where $D$ is the semi-circular disk in the upper half-plane defined by $x^2 + y^2 \le 4$ and $y \ge 0$, is:

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### 1. Strategy & Theoretical Principles Convert to polar coordinates $(r, \theta)$: $$x = r \cos\theta, \quad y = r \sin\theta, \quad dx \, dy = r \, dr \, d\theta$$ $$\sqrt{x^2 + y^2} = r$$ For the semi-circular disk in the upper half-plane of radius $R = 2$: - Radial coordinate: $0 \le r \le 2$ - Angular coordinate: $0 \le \theta \le \pi$ ### 2. Analytical Formulation & Calculation $$I = \int_0^\pi \int_0^2 (r) (r \, dr) \, d\theta = \int_0^\pi d\theta \int_0^2 r^2 \, dr$$ $$I = [\theta]_0^\pi \left[ \frac{r^3}{3} \right]_0^2 = \pi \left(\frac{8}{3}\right) = \frac{8\pi}{3}$$ Evaluating numerically: $$I = \frac{8 \times 3.14159}{3} = \frac{25.1327}{3} \approx 8.3776 \approx 8.38$$ ### 3. Distractor Forensics - Forgetting the Jacobian factor $r$ gives $\int r dr = [r^2/2]_0^2 = 2 \implies 2\pi \approx 6.28$. - Integrating over the full circle gives $16\pi/3 \approx 16.76$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always include the polar area element Jacobian: $dA = r \, dr \, d\theta$. - **Standard Reference:** Thomas, G.B., *Calculus*, Chapter 15.

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