- #1

- 115

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter wifi
- Start date

- #1

- 115

- 1

- #2

Dick

Science Advisor

Homework Helper

- 26,263

- 619

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?

Well, what does Stoke's theorem say? How do you think you might check it? I kind of think you might want to compute a line integral and see if you get the same thing as when you compute a surface integral, yes?

- #3

- 115

- 1

Thanks Dick. I computed the surface integral & I got 101/6. Is this correct?

- #4

Dick

Science Advisor

Homework Helper

- 26,263

- 619

Thanks Dick. I computed the surface integral & I got 101/6. Is this correct?

I have no idea if you are correct. You just asked how to start it. How did you get that so fast? It doesn't look that easy to me. Certainly not this late. Can you show your method?

- #5

SteamKing

Staff Emeritus

Science Advisor

Homework Helper

- 12,798

- 1,670

- #6

- 115

- 1

SteamKing, the region is the fourth quadrant. Positive y is pointing vertically upwards. This is just the way the problem was given.

According to Stokes' Theorem, ## \int\limits_S \, (\nabla \times \vec{v}) \cdot d\vec{a} = \oint\limits_P \vec{v} \cdot d\vec{\ell} ##.

For the RHS I did the following:

##y=2x-2 \Rightarrow dy=2dx ##

## \vec{v} \cdot d\vec{\ell}=(x-3)^3 \hat{x} + (x-y)^3 \hat{y} ##

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 ##.

According to Stokes' Theorem, ## \int\limits_S \, (\nabla \times \vec{v}) \cdot d\vec{a} = \oint\limits_P \vec{v} \cdot d\vec{\ell} ##.

For the RHS I did the following:

##y=2x-2 \Rightarrow dy=2dx ##

## \vec{v} \cdot d\vec{\ell}=(x-3)^3 \hat{x} + (x-y)^3 \hat{y} ##

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 ##.

Last edited:

- #7

- 115

- 1

Anyone?

- #8

CAF123

Gold Member

- 2,950

- 88

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 ##.

The path integral should be split into three parts: One along the x axis, one along the y axis and one along the line y=2x-2. (I think) you have only considered the path along the line y=2x-2 in (1). (2) does not make sense because the LHS is a scalar and the RHS is a vector.

I obtain a different answer than you do, but a negative is ok since the integral is a measure of the net flux through a surface. +/- depending on whether there is a net outward or net inward flux respectively.

- #9

- 115

- 1

I skipped the other 2 integrals because I believe they come out to be 0.

- #10

- 115

- 1

- #11

CAF123

Gold Member

- 2,950

- 88

I skipped the other 2 integrals because I believe they come out to be 0.

Can you show your work?

- #12

- 115

- 1

Certainly! I'll type it out right now. Thanks for the help by the way!

- #13

- 115

- 1

## \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 = 6 \int \int (x-(2x-2))^2dx \ dy = 6 \int \int (2-x)^2dx \ dy = 6 \int \int (x^2-4x+4) dx \ dy##.

So putting in the limits & carrying out the integration, is just left for the LHS. Right?

- #14

- 115

- 1

- #15

CAF123

Gold Member

- 2,950

- 88

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##

I think it is correct up to here. Putting y=2x-2 in your next step means that y=2x-2 everywhere on the planar triangular surface, which is clearly not the case. The next step from here is to put in the appropriate limits of integration for x and y. The integration limits for y will depend on x since the region of integration is not rectangular.

- #16

- 115

- 1

So that means I must integrate over y first and then x?

- #17

- 115

- 1

The limits would then be y: 0 -> 2x-2 and x: 1 -> 0?

- #18

CAF123

Gold Member

- 2,950

- 88

So that means I must integrate over y first and then x?

Or x then y, as long as the outer integral has only constants. Integrating wrt to y then x or x then y will result in different limits of integration but the end result is of course the same.

Your limits in post #17 are correct.

- #19

- 115

- 1

- #20

CAF123

Gold Member

- 2,950

- 88

The line integral around the boundary of the surface is counterclockwise, so the oreintation of the surface is already known. So yes, you are right, the limits for x are from 1 to 0.

- #21

- 115

- 1

Share: