- #1
- 115
- 1
Problem:
Check Stokes's theorem for the vector function ## v=(x-y)^3 \hat{x} + (x-y)^3 \hat{y} ## using the area and perimeter shown below.
Where do I start?
Thanks Dick. I computed the surface integral & I got 101/6. Is this correct?
For the RHS I did the following:
##y=2x-2 \Rightarrow dy=2dx ## (1)
## \vec{v} \cdot d\vec{\ell}=(x-3)^3 \hat{x} + (x-y)^3 \hat{y} ##(2)
Is this an acceptable answer? Does it make sense that the answer is negative?
## \int \vec{v} \cdot d\vec{\ell} = \int\limits_0^1 (x-3)^3dx + \int\limits_0^1 (x-(2x-2))^32dx = -20 ##.
I skipped the other 2 integrals because I believe they come out to be 0.
So for the LHS of Stokes's theorem, I get:
## \nabla \times \vec{v} = (\frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y})\hat{z} = 6(x-y)^2 \hat{z}## (since ## \vec{v} ## does not have a z-component)
## (\nabla \times \vec{v}) \cdot d\vec{a} = [6(x-y)^2 \hat{z}] \cdot [dx \ dy \ \hat{z}] = 6(x-y)^2dx \ dy##
## \int (\nabla \times \vec{v}) \cdot d\vec{a} = \int \int 6(x-y)^2dx \ dy##
So that means I must integrate over y first and then x?
I integrated wrt to y & then x & got an answer of 7. I wasn't sure if the x limits should be from 0->1 or 1->0. I picked 1->0 because that follows the counter-clockwise path, is that correct?