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.

The given integral is:

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.

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 $0 \le z \le 1$.

Substitute the cylindrical coordinates into the integral:

Now the integral becomes:

Finally, the integral becomes:

The final answer is $2\pi$.