Why am I getting half the correct answer by using Green's Theorem?

Question

I had this homework problem that asked me to use Green's Theorem to solve it, so I did. Unfortunately, my answer was wrong. I looked for an error in my reasoning, but did not find it. I eventually solved by way of the line integral, which is usually harder than using Green's Theorem, so now I know that the correct answer is $-64\pi$. Yet, the answer I get from trying to use Green's Theorem is $-32\pi$, exactly half. Below is my attempt, and I'd appreciate a pointer as to where my error is.

$\vec{F} = 4y\hat{i} + 5xy\hat{j}$

$C =$ circle of radius $4$ centered on the origin, oriented counter-clockwise.

$$\begin{align*} \int_C \vec{F}\cdot\text{d}\vec{r} &= \int^{2\pi}_0\int^4_0 (5r\sin(\theta)-4) \text{ d}r\text{ d}\theta \\ &= \int^{2\pi}_0 \left.\left( \frac{5}{2}r^2\sin(\theta) - 4r \right)\right|^{r=4}_{r=0} \text{ d}\theta \\ &= \int^{2\pi}_0 (40\sin(\theta)-16) \text{ d}\theta \\ &= \left. (-40\cos(\theta)-16\theta) \right|^{\theta=2\pi}_{\theta=0} \\ &= -32\pi \\ &\ne -64\pi \end{align*}$$

score 5 · Accepted Answer · answered Oct 30 '13 at 16:53

5

When you switch to polar coordinates, you should replace $dx dy$ with $r dr d\theta$. You are missing a factor of $r$.

answered Oct 30 '13 at 16:53

abnry

14,664

D'oh! That's exactly it! In my defense, the last time I had to worry about that was more than two years ago... >_> – El'endia Starman Oct 30 '13 at 16:57
Yup, the larger $r$ is, the bigger of a sector $drd\theta$ will be. That's why you have to introduce a factor of $r$. – abnry Oct 30 '13 at 18:17
That, and I remember it's the Jacobian for that transformation. – El'endia Starman Oct 30 '13 at 18:24

Why am I getting half the correct answer by using Green's Theorem?

1 Answers1