# Thread: Using Gauss-Green formula to evaluate a path integral

1. ## Using Gauss-Green formula to evaluate a path integral

So the curve C is a triangle with base of 2, height of 2 and is parameterized by {x[t],y[t]} in the counterclockwise direction.

Use the Gauss-Green formula to evaluate the path integral:
*Integral C* [E^(-x^2)+x y^2]dx + [x^3+ sqroot(1+y^3)]dy

I'm completely stuck because I'm not sure how to get all the x's and y's in terms of t, which should be the first step I'm guessing.

I have many problems to do just that are similar to this one. Could I please get a step-by-step so I can apply to others? That would be awesome

2. Originally Posted by Lucky1
So the curve C is a triangle with base of 2, height of 2 and is parameterized by {x[t],y[t]} in the counterclockwise direction.

Use the Gauss-Green formula to evaluate the path integral:
*Integral C* [E^(-x^2)+x y^2]dx + [x^3+ sqroot(1+y^3)]dy

I'm completely stuck because I'm not sure how to get all the x's and y's in terms of t, which should be the first step I'm guessing.

I have many problems to do just that are similar to this one. Could I please get a step-by-step so I can apply to others? That would be awesome
What you're trying to evaluate is

$\int_C \left(e^{-x^2} + xy^2\right)\,dx +\left( x^3 + \sqrt{1+y^3}\right)\, dy$ where C is the ccw path around the triangle. However by Green's theorem in the plane

$\int_C P\,dx + Q\,dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = \int_0^2 \int_0^x \left( 3x^2 - 2xy\right)\, dy\,dx$ much easier to do!

3. I understand that the path integral would be much easier. However, the problem calls for me to use the Gauss-Green formula, which sucks.

Thanks for looking tho!

4. Originally Posted by Lucky1
I understand that the path integral would be much easier. However, the problem calls for me to use the Gauss-Green formula, which sucks.

Thanks for looking tho!
He just told you that the area integral is easier to do. And that is using the Gauss-Green formula!

5. Oooh! Sorry! They even gave me the area formula...I just didn't understand the question correctly. Thinking too much into it....Thanks!