# Green's Theorem and converting to polar co-ords

• May 6th 2009, 11:24 AM
chella182
[SOLVED] Green's Theorem and converting to polar co-ords
The question goes...

Use Green's Theorem to evaluate $\oint_{\Gamma}(y^3dx-x^3dy)$, where $\Gamma$, mapped counter-clockwise, is the circle $x^2+y^2=4$. [Hint: you may find it convenient to introduce the transformation to polars.]

Letting $x=\cos{\theta}$ and $y=\sin{\theta}$ and differentiating et cetera, I end up with...
$-\int_{0}^{2\pi}\sin^{4}{\theta}+\cos^{4}{\theta} d\theta$
And I'm stuck from here. I'm guessing it's something to do with relationship $\cos^{2}{\theta}+\sin^{2}{\theta}=1$ somewhere, but I'm just not getting it. Can anyone help?
• May 6th 2009, 11:44 AM
NonCommAlg
Quote:

Originally Posted by chella182
The question goes...

Use Green's Theorem to evaluate $\oint_{\Gamma}(y^3dx-x^3dy)$, where $\Gamma$, mapped counter-clockwise, is the circle $x^2+y^2=4$. [Hint: you may find it convenient to introduce the transformation to polars.]

Letting $x=\cos{\theta}$ and $y=\sin{\theta}$ and differentiating et cetera, I end up with...
$-\int_{0}^{2\pi}\sin^{4}{\theta}+\cos^{4}{\theta} d\theta$
And I'm stuck from here. I'm guessing it's something to do with relationship $\cos^{2}{\theta}+\sin^{2}{\theta}=1$ somewhere, but I'm just not getting it. Can anyone help?

you didn't use Green's Theorem! first use the theorem and then change to polar. you'll end up with a very simple integral.
• May 6th 2009, 11:55 AM
chella182
So doing the whole P and Q thing I end up with $-3\iint_{A}x^2+y^2 dA$, and NOW do I sub in $\cos{\theta}$ and $\sin{\theta}$ for $x$ and $y$?
• May 6th 2009, 12:00 PM
NonCommAlg
Quote:

Originally Posted by chella182
Oh I thought we were meant to change to polars first o.O

well, what you did wasn't even changing to polar. it was parametrizing the unit circle and not even the given circle. you can't use polar for line integrals. it's used for double integrals only!

in polar you put $x=r \cos \theta, \ y=r \sin \theta$ not $x=\cos \theta, \ y= \sin \theta.$ (you have two variables $r , \theta.$)
• May 6th 2009, 12:02 PM
chella182
Okay so what I edited my post you quoted to is wrong as well then?
• May 6th 2009, 12:07 PM
NonCommAlg
Quote:

Originally Posted by chella182
So doing the whole P and Q thing I end up with $-3\iint_{A}x^2+y^2 dA$, and NOW do I sub in $\cos{\theta}$ and $\sin{\theta}$ for $x$ and $y$?

this is correct! now put $x=r \cos \theta, \ y=r \sin \theta.$ the Jacobian in polar coordinates is $r$ and $x^2+y^2=r^2.$ thus: $-3\iint_{A}x^2+y^2 dA=-3 \int_0^{2 \pi} \int_0^2 r^3 dr d \theta,$ which is easy to evaluate.
• May 6th 2009, 12:13 PM
chella182
The $r^3$ comes from the $r$ in the $rdrd\theta$ mutliplied by the $r^2$ fromthe circle yes? So it'd be...

$\int_{0}^{2\pi}[\frac{r^4}{4}]_{0}^{2} d\theta$

$=\int_{0}^{2\pi}4 d\theta$

$=[4\theta]^{2\pi}_{0}$

$=8\pi$

That right?
• May 6th 2009, 12:19 PM
NonCommAlg
Quote:

Originally Posted by chella182
The $r^3$ comes from the $r$ in the $rdrd\theta$ mutliplied by the $r^2$ fromthe circle yes? So it'd be...

$\int_{0}^{2\pi}[\frac{r^4}{4}]_{0}^{2} d\theta$

$=\int_{0}^{2\pi}4 d\theta$

$=[4\theta]^{2\pi}_{0}$

$=8\pi$

That right?

yest it is correct but don't forget that you have a factor $-3$ outside the integral. so the final answer is $-24 \pi.$
• May 6th 2009, 12:21 PM
chella182
Oh sugar, yeah, thanks for that :)