Greens Theorem Area of ellipse

In summary: The final answer should be (1/2) a^2 sin2t0, since the cosine term will cancel out in the integration.
  • #1
gtfitzpatrick
379
0

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]
 
Physics news on Phys.org
  • #2
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 ...
 
  • #3
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]
...
 
  • #4
.. 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
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.
 
  • #6
Your original result is actually correct, but you made a sign error in your work.
 

Related to Greens Theorem Area of ellipse

1. What is Green's Theorem and how is it related to the area of an ellipse?

Green's Theorem is a fundamental theorem in multivariable calculus that relates the area of a two-dimensional region to the line integral along the boundary of that region. It is specifically useful for calculating the area of an ellipse because it can be expressed as a double integral, making the calculation simpler.

2. How does Green's Theorem help in finding the area of an ellipse?

Green's Theorem allows us to convert a line integral into a double integral, which is easier to calculate. This allows us to find the area of an ellipse by integrating over the entire region instead of just the boundary. This method is particularly useful when the boundary of the ellipse is difficult to describe or when the ellipse is rotated.

3. Is Green's Theorem only applicable to ellipses?

No, Green's Theorem can be applied to any two-dimensional region with a smooth boundary. It is commonly used for other shapes such as circles, rectangles, and irregular polygons.

4. Can Green's Theorem be used to find the area of a three-dimensional shape?

No, Green's Theorem is only applicable to two-dimensional regions. To find the area of a three-dimensional shape, we would need to use other methods such as surface integrals or volume integrals.

5. Are there any limitations to using Green's Theorem to find the area of an ellipse?

Green's Theorem can only be used for finding the area of an ellipse if the ellipse is a simple closed curve, meaning it does not intersect itself. If the ellipse has a self-intersecting boundary, Green's Theorem cannot be applied and other methods must be used to find the area.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
882
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
669
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
709
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
305
  • Calculus and Beyond Homework Help
Replies
16
Views
842
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
453
Back
Top