Green's Theorem (integration problem, not concept)

In summary: Then take the square root of both sides to get x = 8 and y = 16. I'm sorry if that's too easy or not easy enough.In summary, the area of the curve C is 2 \oint_{0}^{2 \pi} (8*cos(t) + 16*sin(t))dt.
  • #1
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So a previous problem said to show that the area of some simple closed curve C was:

[tex]A= \frac{1}{2} \oint_{C} (xdy - ydx)[/tex]

Simple enough. My problem says to find the area of the curve

[tex]x^{2/3} + y^{2/3} = 4[/tex]

using that formula.

So off to polar coordinate land I go. r = 4, x = cos(t), dx = -sin(t), y = sin(t), dy = cos(t), right?

So after plugging everything in and taking out the r, I get:

[tex]A = 2 \oint_{0}^{2 \pi} (cos(t)^{5/3} + sin(t)^{5/3})dt[/tex]

Correct?

How do I integrate that? They aren't squared. And I can't find any trig formula to simplify that or any integration formula. Do I have to venture into infinite series?
 
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  • #2
I don't think you're parametrizing your curve correctly--one way to check is to substitute a point from your parametrized expression back into the original expression. If you let (x(t), y(t)) = (sin^3 t, cos^3 t), how might you tweak that?
 
  • #3
Now I'm completely lost. I've been doing this assignment and another lab since yesterday morning, so my mind is fried.

I think I understand that r =/= 4 (although I am having a brain fart and can't figure out what it should be), but I still don't understand your last sentence.
 
  • #4
You are trying to find a parametric representation of the curve [tex]x^{2/3} + y^{2/3} = 4[/tex]. That means you want a function c(t) = (x(t), y(t)) such that for all t, [tex]x(t)^{2/3} + y(t)^{2/3} = 4[/tex], and also you want every point (x, y) on the curve to be equal to (x(t), y(t)) for some t. Are you with me? I'm saying you could try choosing c(t) = (cos^3(t), sin^3(t)) and see how you can alter it so that it's right.
 
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  • #5
Yeah, I ended up doing that but didn't know how to integrate sin^2(t)cos^2(t). I really hate that. I understand the concept but I don't know how to integrate. =/ It was like 1 - cos(4t) or something, and integrating that is a snap, but my trig sucks. :(
 
  • #6
If you wanted to simplify the curve eqn you could cube the whole equation to get x^2 + y^2 = 64.
 

1. What is Green's Theorem and how is it used in integration problems?

Green's Theorem is a mathematical tool used to solve line integrals over a closed curve in the xy-plane. It relates the line integral to a double integral over the region enclosed by the curve, making it easier to evaluate complex integrals.

2. What are the conditions for using Green's Theorem?

In order to use Green's Theorem, the region enclosed by the curve must be simply connected, meaning there are no holes or self-intersections. The curve must also be smooth, and the functions involved must have continuous partial derivatives within the region.

3. How do I know if I should use Green's Theorem in a given integration problem?

If you are given a line integral over a closed curve in the xy-plane, and the region enclosed by the curve satisfies the conditions for Green's Theorem, then you should use it to simplify the integration. Additionally, if you are struggling to evaluate a complex line integral using traditional methods, Green's Theorem may provide a more straightforward solution.

4. Can Green's Theorem be applied to regions in three-dimensional space?

No, Green's Theorem is specifically designed for regions in the xy-plane. For regions in three-dimensional space, the appropriate tool is Stokes' Theorem.

5. Are there any common mistakes when using Green's Theorem?

One common mistake when using Green's Theorem is forgetting to change the orientation of the curve in the double integral. This results in a negative sign in the final calculation. It is also important to check that the conditions for Green's Theorem are satisfied before using it in an integration problem.

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