Greens Theorem Area of ellipse

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Homework Help Overview

The discussion revolves around finding the area swept out by a line from the origin to an ellipse defined parametrically by x=acos(t) and y=asin(t), as the parameter t varies from 0 to a constant t_{0} between 0 and 2\pi. The problem involves applying Green's Theorem in reverse.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Green's Theorem and the integration process involved in calculating the area. There are questions about the correctness of certain trigonometric identities and the implications of the variable t_{0} on the area calculation.

Discussion Status

Some participants have pointed out potential mistakes in the original poster's calculations, particularly regarding trigonometric identities and sign errors. There is a suggestion that the range of t_{0} does not affect the equations significantly, while others note that the area should relate to the total area of the ellipse.

Contextual Notes

Participants are considering the implications of t_{0} varying from 0 to 2\pi, and there is a discussion about the nature of the ellipse being a specific case that resembles a circle.

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

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