Greens Theorem Area of ellipse

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gtfitzpatrick
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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 [itex]t_{0}[/itex] where [itex]t_{0}[/itex] is a constant between 0 and 2[itex]\pi[/itex]

Homework Equations


The Attempt at a Solution



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

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

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

[itex]-\frac{1}{2}\int^{t_0}_{0} a(1) dt[/itex]

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

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 [itex]t_{0}[/itex] where [itex]t_{0}[/itex] is a constant between 0 and 2[itex]\pi[/itex]

Homework Equations


The Attempt at a Solution



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

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

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

[itex]-\frac{1}{2}\int^{t_0}_{0} a(1) dt[/itex]

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

Are you sure? I don't think sin^2 (t) - cos^2(t) is equal to -1 ...
 
ahh yes, what a mistake a ta make a

after i integrate i should get -sin2t[itex]_0[/itex] but my problem is still how to use the information that [itex]t_o[/itex] varies from 0 to 2[itex]\pi[/itex]
...
 
.. 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
 
gtfitzpatrick said:
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.