# Green's Theorem

• February 2nd 2011, 11:37 AM
ramdrop
Green's Theorem
Hey,

I have a tutorial question that is starting to bug me.
I know the formula is:

A = $\frac{1}{2}\oint_{C}(x\,\mathrm{d}y + y\, \mathrm{d}x)$
I have the problem:
Using Green's theorem, find the area of the disc of radius a.

I have notes and a previous example, but unsure where to start, Do I use the fact that:
$x^2+y^2=r^2$ where r =1

cheers
• February 2nd 2011, 11:39 AM
TheEmptySet
Quote:

Originally Posted by ramdrop
Hey,

I have a tutorial question that is starting to bug me.
I know the formula is:

A = $\frac{1}{2}\oint_{C}(x\,\mathrm{d}y + y\, \mathrm{d}x)$
I have the problem:
Using Green's theorem, find the area of the disc of radius a.

I have notes and a previous example, but unsure where to start, Do I use the fact that:
$x^2+y^2=r^2$ where r =1

cheers

In polar coordinates with $r=1$ you get

$x=\cos(\theta) \implies dx=-\sin(\theta)d\theta$ and
$y=\sin(\theta) \implies dy=\cos(\theta)d\theta$

Now $\theta=0..2\pi$
Can you finish from here?
• February 2nd 2011, 11:41 AM
FernandoRevilla
Use the fact that:

$C \equiv\begin{Bmatrix} x=a\cos t\\y=a\sin t\end{matrix}\quad(t\in [0,2\pi])$

Fernando Revilla

Edited: Sorry, I did't see TheEmptySet's post.
• February 2nd 2011, 12:32 PM
ramdrop
ah okay, I was right in assuming that, cheers guys
• February 2nd 2011, 01:03 PM
ramdrop
I got a final answer of 0..

having replacing:
$x = cos{theta}, y = sin(theta), x1 = -sin(theta), y1 = cos(theta)$

then subbing it in I got:
$cos^2(theta) - sin^2(theta)$ with the limits 2pi and 0.

I then changed this to $cos(2*theta)$ by the formula and integrated to get:
$0.25sin(2*theta)$

and then finally putting the limits in, i got 0, surely it shouldnt be 0..
• February 2nd 2011, 01:19 PM
TheEmptySet
Quote:

Originally Posted by ramdrop
I got a final answer of 0..

having replacing:
$x = cos{theta}, y = sin(theta), x1 = -sin(theta), y1 = cos(theta)$

then subbing it in I got:
$cos^2(theta) - sin^2(theta)$ with the limits 2pi and 0.

I then changed this to $cos(2*theta)$ by the formula and integrated to get:
$0.25sin(2*theta)$

and then finally putting the limits in, i got 0, surely it shouldnt be 0..

I didn't notice at first but you have a typo in your formula it should be

$\displaystyle \frac{1}{2}\oint xdy-ydx$

You need to change the sign on the sine term (haha pun intended!) then you will get the answer you wish
• February 2nd 2011, 01:33 PM
ramdrop
So stupid, i copied it from my notes and i must have wrote it down wrong, because i haad both a + and a -, so I took a guess... my bad

I got pi now :)