Find suitable curve, Green's theorem

[usn7564]

Homework Statement

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.

 

[LCKurtz]

Homework Statement

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 $$1 \leq \frac{x^2}{4}+y^2$$

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

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 $$0 \leq \theta \pi$$ $\color{red}{2\pi}$? and $$0 \leq r \leq 1$$.

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

 

[SammyS]

...

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

Why is the integrand r(1 - r³) ?

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

 

[LCKurtz]

Why is the integrand r(1 - r³) ?

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

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

 

[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.