# Test problem

1. Nov 30, 2007

### amolv06

I'm posting this problem because, one I found it fun, and two because I have a couple of questions about it. You start with the equation of a circle

$$x^{2}+y^{2}=16$$.

Then you are told that another circle exists with $$r=1$$. This circle always touches the edge of the circle with r=4, and rolls along its edge. We are told that we're supposed to track a point on the edge of the circle with r=1. The path the point makes is given by

$$\vec{r}(t)=(5cos(t)-cos(5t))\hat{i}+(5sin(t)-sin(5t))\hat{j}$$ where $$0\leq t \leq2\pi$$.

We're supposed to find the area bound by $$\vec{r}$$. We're given the vector field $$\vec{w}=<0,x>$$. We're supposed to find the area bound by r using Green's theorem. Now, I think I figured out how to do it properly. Since

$$\frac{\delta Q}{\delta x}-\frac{\delta P}{\delta Y} = 1$$

I believe the equation should reduce too

$$\oint_{c} x dy = \int^{2 \pi}_{0}[(5cos(t)-cos(5t))][(5cos(t)-5cos(5t))]dt=30 \pi$$.

This is what I put down on the test, anyway. Hopefully I got it right. I wasn't entirely convinced, however. I thought there might be another, even easier way that I could try. I figured if I could just get the area of an eighth of the curve, from when the point touches the r=4 circle at t=0 to where it touches at

$$t=\frac{\pi}{4}$$

I would be able to multiply that area by 8 and comfortably on with my day. While this wasn't a perfect circle, I figured I could set up the limits of integration right and get an answer. I tried it as such:

$$\int^{\frac{\pi}{4}}_{0} \int^{4+2sin(2 \theta)}_{0} r dr d\theta$$

As I'm sure it's needless to say, this lead me to unexpected results. Am I just being stupid, or do I lack understanding of integration with polar coordinates, or is it just some small mistake? I was just wondering where I went wrong. After everything, what I'm getting is

$$8 \int^{\frac{\pi}{4}}_{0} \int^{4+2sin(2 \theta)}_{0} r dr d\theta = 18 \pi + 32$$