Paper Homework 7

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 $y\leq \sqrt{2x-x^2}$.