Homework 7 was due on Tuesday 25th November 2014. Questions 1, 2(c), 3(b), 5(b) and 9 were graded.
Question 1. I don't have much to say about this one, most people got it right or perhaps got two parts the wrong way round.
Question 2(c). The most common error in this question—which I understand was also a common mistake on the WebWork—was writing $\ln(x)$ as the antiderivative of $\frac{1}{x}$, instead of $\ln|x|$. The absolute value matters here, because the domain of your function and the vector field have to be the same: $\frac{1}{x}$ is defined for all $x \ne 0$ but $\ln(x)$ is only defined for $x>0$.
Question 3(b). This was done well by most people who parametrised the half-circle correctly. The most common error in parametrising the circle was using $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ instead of $\langle 2\cos t, 2\sin t \rangle$. We need the $2$ there because the half-circle has radius $2$.
Question 5(b). Most people had the right idea with this one: use $\mathbf{r}(t) = \langle \cos t, t \rangle$ and let $t$ range between $0$ and $2\pi$. However, most people had their limits going in the wrong direction. Because the path starts at $(1,2\pi)$, i.e. when $t=2\pi$, and ends at $(1,0)$, i.e. when $t=0$, your integral should look like $\int_{2\pi}^0$ and not $\int_0^{2\pi}$. Be very careful with making sure your paths are being parametrised in the right direction. If it helps, write something like $2\pi \ge t \ge 0$ instead of $0 \le t \le 2\pi$... even though these 'mean the same thing', they convey different directions of $t$.
Question 9. If you tried to do this question without using the fact that the vector field is conservative, you probably screwed it up. Those that did notice that the vector field was conservative fell into two categories. The first category used the fundamental theorem of line integrals... they found a potential function and plugged in the end points, and usually got the correct answer without too much hassle. The second category used the fact the integral is independent of the path, and instead integrated over a straight line segment... this amounted to a lot more work, but people still usually got the correct answer.