# Homework Help: Greens Theorem Area of ellipse

1. Jul 21, 2012

### gtfitzpatrick

1. The problem statement, all variables and given/known data

Find the area swept out by the line from the origin to the ellipse x=acos(t) y=asin(t) as t varies from 0 to $t_{0}$ where $t_{0}$ is a constant between 0 and 2$\pi$

2. Relevant equations

3. The attempt at a solution

so using Greens Theorem in reverse i get A=$\frac{1}{2}\oint_{c} ydx-xdy$

x=acos(t) dx=-asin(t)
y=asin(t) dy=cos(t)

so sub into my equation i get $\frac{1}{2}\int^{t_0}_{0} a(sin^2 (t) - cos^2(t)) dt$

$-\frac{1}{2}\int^{t_0}_{0} a(1) dt$

I think im good up to here, i then integrate and get $-\frac{1}{2} at_0$ but im not sure about how to use the information that $t_0$ varies from 0 to 2$\pi$

2. Jul 21, 2012

### w4k4b4lool4

Are you sure? I don't think sin^2 (t) - cos^2(t) is equal to -1 ...

3. Jul 21, 2012

### gtfitzpatrick

ahh yes, what a mistake a ta make a

after i integrate i should get -sin2t$_0$ but my problem is still how to use the information that $t_o$ varies from 0 to 2$\pi$
...

4. Jul 21, 2012

### w4k4b4lool4

.. I don't think you have to worry about the range of t_0. All this means is that the total area will be less or equal to that of the ellipse, it doesn't change the equations.
Wakabaloola

5. Jul 21, 2012

### uart

Are you sure that's the problem statement. It's not a very general ellipse, seeing as it is a circle. In this case of course the area is simply that of a sector of angle t0, hence A = (1/2) a^2 t0.

6. Jul 21, 2012

### vela

Staff Emeritus
Your original result is actually correct, but you made a sign error in your work.

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