Greens Theorem Area of ellipse

[gtfitzpatrick]

Homework Statement

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

Homework Equations

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

 

[w4k4b4lool4]

Homework Statement

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

Homework Equations

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

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

 

[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
...

 

[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

 

[uart]

ellipse x=acos(t) y=asin(t)

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.

 

[vela]

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