# Find suitable curve, Green's theorem

1. May 3, 2013

### usn7564

1. The problem statement, all variables and given/known data
Excuse my terminology, not sure what the actual translations are.

Find a simple (no holes in it), closed, positively oriented, continuously differentiable curve T in the plane such that:

$$\int_{T}(4y^3+y^2x-4y)dx + (8x +x^2y-x^3)dy$$

is as big as possible, finally calculate the value of the integral.

The attempt at a solution
I used Green's Theorem to get a rather simple (or so I figured) expression:
$$12\iint_{D}1-(\frac{x^2}{4}+y^2)dxdy$$
This made it clear (I presume) that I wanted the ellipse satisfying the equation $$1 \leq \frac{x^2}{4}+y^2$$
Then I parametrised it with
$$x = 2cos\theta$$
$$y = sin\theta$$

And got
$$12 \cdot 2\iint r(1-r^3)drd\theta$$

With the 2 and extra r due to the Jacobian determinant. Solved it with $$0 \leq \theta \pi$$ and $$0 \leq r \leq 1$$.

I get 6pi, done it numerous times and it stays at 6pi. The answer should be 12pi. Any idea where I'm messing up? I haven't really parametrized an ellipse for a while but I don't remember there being any oddities there, though I guess that's an area where I could have messed up. Quite google search lead me to believe it's not it though, so really not sure.

2. May 3, 2013

### LCKurtz

No. You want the region where the integrand is positive, which is the inside of that ellipse, not the exterior as you have written.

Apparently you did integrate over the interior of the ellipse. That integral is correct and when I evaluate it I get $12\pi$.

3. May 3, 2013

### SammyS

Staff Emeritus
Why is the integrand r(1 - r3) ?

Shouldn't it be r(1 - r2), or am I overlooking something?

4. May 3, 2013

### LCKurtz

You are right, I missed that when I read his proof. The integrand should be $r-r^3$, which is how you get the $12\pi$.

5. May 3, 2013

### usn7564

God, I need to proof read next time I make a topic. It was $$r-r^3$$ though I can find my error I believe. I integrated $$0 \leq \theta \leq \pi$$ thinking that's the part of the integral that will give me a positive value. Come to think of it that makes less sense when I'm not doing a regular single variable integral, how silly of me. Just slipped my mind that I was working with the domain and not 'everything' so to speak.

Thanks a lot. Won't make that mistake again.