# Evaluate the integral with C be the boundary of the domain

1. May 7, 2013

### tavo0116

1. The problem statement, all variables and given/known data

Let C be the boundary of the domain enclosed between y = x^2 and y = x. Assuming C is oriented counterclockwise. Evaluate the integral
∫c (6xy+e^(-x^2))dx

2. Relevant equations

I was thinking of using Green's Theorem. Would be the approach be correct?

3. The attempt at a solution

d/dy (6xy+e^(-x^2))dx = 6x, and integrate that I got 3x^2
What would be the limit? Would the problem be double integral since there is no "dy".
Would the limit be from 0 to 1 since y = x^2 is a parabola, and y = x is a diagonal line through the origin?

2. May 7, 2013

### LCKurtz

One side of Green's theorem is a double integral. If you were asked to find the area between those two curves, could you set up the double integral with limits for that? That is what you want, with the integrand being 6x.

Last edited: May 7, 2013
3. May 7, 2013

### tavo0116

But there is no "dy" , I still don't understand that.
Would the limit of x is from 0 to 1, and limit of y is from y to √y ?

4. May 7, 2013

### LCKurtz

Your original problem is$$\int_C (6xy + e^{-x^2})dx + 0dy$$Does that help? Write down the other side of Green's theorem.

Last edited: May 7, 2013
5. May 7, 2013

### tavo0116

d/dx (0) - d/dy (6xy+e^(-x^2))
∫0 to 1 ∫0 to 1 (-6x)

I've tried to find the limits of x by setting up x = x^2, which then gives me x = 0, and x = 1, and for y, I use √y=y, which gives me y = 0, and y = 1

Would those be the right limits for the double integral?

6. May 7, 2013

### LCKurtz

No. 0 to 1 on both integrals describes a square. Your region isn't a square. Look in your calculus book for the formula for the area between two curves.