Evaluate the triple integral of (x+y)2 over the region bounded by x=0, x=2, y=0, y=sqrt(4-x2), z=0, z=1.

Step 1: Identify the integral and its limits. The given integral is: $\int_{0}^{1} \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} (x^2 + y^2) \, dy \, dx \, dz$

Step 2: Analyze the region of integration in the $xy$-plane. The limits for $y$ are $0 \le y \le \sqrt{4-x^2}$, which implies $y^2 \le 4-x^2$ or $x^2+y^2 \le 4$ (since $y \ge 0$). The limits for $x$ are $0 \le x \le 2$. Combined, these limits describe the quarter-circle of radius $2$ in the first quadrant of the $xy$-plane.

Step 3: Convert to cylindrical coordinates. For the $xy$-plane, we use cylindrical coordinates: $x = r \cos \theta$ $y = r \sin \theta$ $x^2 + y^2 = r^2$ The differential $dy \, dx$ becomes $r \, dr \, d\theta$.

The region $0 \le x \le 2$, $0 \le y \le \sqrt{4-x^2}$ in the $xy$-plane corresponds to: $0 \le r \le 2$ (radius from origin to the circle $x^2+y^2=4$) $0 \le \theta \le \frac{\pi}{2}$ (first quadrant)

The limits for $z$ remain $0 \le z \le 1$.

Step 4: Rewrite the integral in cylindrical coordinates. Substitute the cylindrical coordinates into the integral: $\int_{0}^{1} \int_{0}^{\pi/2} \int_{0}^{2} (r^2) \, r \, dr \, d\theta \, dz$ $\int_{0}^{1} \int_{0}^{\pi/2} \int_{0}^{2} r^3 \, dr \, d\theta \, dz$

Step 5: Evaluate the innermost integral with respect to $r$. $\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{2}$ $= \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4$

Step 6: Evaluate the middle integral with respect to $\theta$. Now the integral becomes: $\int_{0}^{1} \int_{0}^{\pi/2} 4 \, d\theta \, dz$ $\int_{0}^{\pi/2} 4 \, d\theta = \left[ 4\theta \right]_{0}^{\pi/2}$ $= 4 \left(\frac{\pi}{2}\right) - 4(0) = 2\pi$

Step 7: Evaluate the outermost integral with respect to $z$. Finally, the integral becomes: $\int_{0}^{1} 2\pi \, dz$ $= \left[ 2\pi z \right]_{0}^{1}$ $= 2\pi (1) - 2\pi (0) = 2\pi$

The final answer is $\boxed{2\pi}$.