Verification of Green's Theorem

[sunnyday11]

Homework Statement

Verify Green's Theorem for F(x,y) = (2xy-x²) i + (x + y²) j
and the region R which is bounded by the curves y = x² and y² = x

Homework Equations

\int _CF dr = \int\int_R (dF₂/dx - dF₁/dy) dxdy

The Attempt at a Solution

For \int _CF dr ,

r(t) = x i + x² j
r'(t) = i + 2x j

0 to 1 \int (2x³ + x² + 2x⁵)dx

= 2/4 + 1/3 + 2/6
= 7/6



For \int\int_R (dF₂/dx - dF₁/dy) dxdy both integrate from 0 to 1.

\int\int (1 - 2x) dxdy = \int (1- 2x)dx = 1-1 = 0


I think I have made an error somewhere but I could not figure out where.

Thank you very much!

 

[HallsofIvy]

There are several errors. First, you have only integrated half way around the boundary- from (0, 0) to (1, 1) along the curve y= x^2. You still need to integrate from (1, 1) back to (0, 0) along the curve x= y^2.

And the integration over the region bounded by those is NOT
\int_0^1\int_0^1 dydx

That is over the square with boundaries y= 0, y= 1, x= 0, x= 1. To integrate over the given region, if you wish to use the order "dydx" then, yes, x goes form 0 to 1 but for every x, y goes from x^2 to \sqrt{x}:
\int_{x=0}^1\int_{y= x^2}^{\sqrt{x}} dydx
//
Or you can use the order dxdy. Then y goes from 0 to 1 while x goes from y^2 to \sqrt{y}:
\int_{y=0}^1\int_{x= y^2}^{\sqrt{y}} dxdy

 

[ojs]

And also be more precise in your notation. Don't use x for your parameter when that is already a variable in the original vector field. It makes no sense to write r(t) = x i + x^2 j unless x is a function of t but then you write x(t) instead of just x. Good notation is always best to solve a problem.

 

[sunnyday11]

Thank you very much for your help!
I got 1/15 for both parts, so I just subtracted one from the other and got 0 which agrees with the other solution.