Find the area of the region bounded by y=z+2y = \sqrt{z+2}y=z+2, y=0y = 0y=0, z=−1z = -1z=−1, and z=2z = 2z=2.
23−2322\sqrt{3} - \frac{2}{3}\sqrt{2}23−322
23(33−1)\frac{2}{3}(3\sqrt{3} - 1)32(33−1)
23(33−2)\frac{2}{3}(3\sqrt{3} - \sqrt{2})32(33−2)
33−13\sqrt{3} - 133−1