**Problem 3: **Almost all students have done well in this problem. The only mistake I have seen is the calculation mistake.

**Problem 7:** A lot of mistakes derived from the wrong picture of the region. It should be the region below the straight line y=3 for 0\leq x\leq \sqrt{6} and the curve y=9-x^2 for \sqrt{6} \leq x\leq 3. Once this is done, you can either write a sum of two integrals \left(\int_0^{\sqrt{6}}\int_0^3+\int_{\sqrt{6}}^3\int_0^{9-x^2}\right)f(x,y)dydx or in a more compact form \int_0^3\int_0^{\min(3, 9-x^2)}f(x,y)dydx.

**Problem 8: **The region of the integral is 0\leq \theta\leq 2\pi, 2\leq r\leq 4 and the integrand is 2\sqrt{16-r^2}. Some students forgot the factor 2 in the integrand.

**Problem 10: **The region of the integral is the upper half of the disc centered at (1,0) with radius 1. In terms of polar coordinates, 0\leq \theta \leq \pi / 2 and 0\leq r\leq 2\cos \theta. The second inequality for r comes from the inequality y\leq \sqrt{2x-x^2}.