# Green's Th. problem

1. Dec 6, 2007

### Rhduke

1. The problem statement, all variables and given/known data
Find area of indicated region: S is bounded by the curves y = 4x and y = 2x^3

note: }a:b{ is what i mean by the definite integral

2. Relevant equations
A(s) = 1/2 }Green's Int{ (-ydx + xdy) = 1/2 }{ }{ (dN/dx - dM/dy)dxdy

3. The attempt at a solution

y' = 4
intersection: 4x = 2x^3 ---> x = sqrt(2)

A(s) = 1/2 }0:sqrt(2){ }4x:4{ (1 + 1) dydx

= 1/2 }0:sqrt(2){ (2(4) - 2(4x)) dx
= 1/2 (8(sqrt(2)) - 4(2))
= 4(sqrt(2)) - 4
(not equal to 8/3)

sorry bare with me i dont totally understand the theorem but i tried

Last edited: Dec 7, 2007
2. Dec 7, 2007

### Defennder

Uh, why is there a need to use Green's theorem to solve this? Couldn't you just find the area by means of a definite integral? I thought Green's theorem only applied to line integrals of vector functions/fields?

3. Dec 7, 2007

### Office_Shredder

Staff Emeritus
I'm not quite sure what you were trying to do (the notation you used is poor, so I have no idea what's being integrated, etc.) but at the top you had the curves as 4x and 2x^2, so the intersection would occur at x=2

4. Dec 7, 2007

### Rhduke

yes u can just integrate the easier way, but the point is to use greens theorem, oops its y = 2x^3

the question wants me to use this equation to solve it:
ok ill use I(a,b) as the integral

I(Greens) (-y/2 dx + x/2 dy) = I I (1/2 + 1/2) dA = A(s)

i guess my real question is how do u find the intervals on which you integrate the double integral

Last edited: Dec 7, 2007
5. Dec 7, 2007

### HallsofIvy

Staff Emeritus
Using Green's theorem, it's easy to show that the area of the region enclosed by the closed path C is $\int_C x dy$ or $\int_C -ydx$.

I also note that y= 4x and y= 2x3 intersect at $(-\sqrt{2},-4\sqrt{2})$, (0, 0), and $(-\sqrt{2},-4\sqrt{2})$ so the region has two symmetric regions, one in the first quadrant and one in the third quadrant. I would recommend finding the area of the one in the first quadrant and doubling.

$\int_C -ydx$ divides into two integrals: On y= 2x3, it is $\int_0^{\sqrt{2}} -2x^3 dx$. On y= 4x, it is $\int_{\sqrt{2}}^0 -4xdx$.