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:

[tex]\int_{T}(4y^3+y^2x-4y)dx + (8x +x^2y-x^3)dy[/tex]

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:
[tex]12\iint_{D}1-(\frac{x^2}{4}+y^2)dxdy[/tex]
This made it clear (I presume) that I wanted the ellipse satisfying the equation [tex]1 \leq \frac{x^2}{4}+y^2[/tex]
Then I parametrised it with
[tex] x = 2cos\theta [/tex]
[tex] y = sin\theta[/tex]

And got
[tex]12 \cdot 2\iint r(1-r^3)drd\theta[/tex]

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

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:

[tex]\int_{T}(4y^3+y^2x-4y)dx + (8x +x^2y-x^3)dy[/tex]

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:
[tex]12\iint_{D}1-(\frac{x^2}{4}+y^2)dxdy[/tex]
This made it clear (I presume) that I wanted the ellipse satisfying the equation [tex]1 \leq \frac{x^2}{4}+y^2[/tex]

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
[tex]12 \cdot 2\iint r(1-r^3)drd\theta[/tex]

With the 2 and extra r due to the Jacobian determinant. Solved it with [tex]0 \leq \theta \pi[/tex] ##\color{red}{2\pi}##? and [tex]0 \leq r \leq 1[/tex].

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
[tex]12 \cdot 2\iint r(1-r^3)drd\theta[/tex]
...

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 [tex]r-r^3[/tex] though I can find my error I believe. I integrated [tex]0 \leq \theta \leq \pi[/tex] 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.