Stokes's Theorem for Vector Function v=(x-y)^3: Area & Perimeter Analysis

[wifi]

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?

 

[Dick]

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?

 

[wifi]

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

 

[Dick]

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?

 

[SteamKing]

Is your region in the first quadrant or the fourth? The convention is to draw positive y pointing up on the page. If your region is in the fourth quadrant, your y values should be negative.

 

[wifi]

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?

$\int \vec{v} \cdot d\vec{\ell} = \int\limits_0^1 (x-3)^3dx + \int\limits_0^1 (x-(2x-2))^32dx = -20$.

 

[wifi]

Anyone?

 

[CAF123]

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.

 

[wifi]

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

 

[wifi]

$\vec{v} \cdot d\vec{\ell}=(x-3)^3 \hat{x} + (x-y)^3 \hat{y}$ should have been$\vec{v} \cdot d\vec{\ell}=(x-3)^3 dx + (x-y)^3 dy$

 

[CAF123]

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

Can you show your work?

 

[wifi]

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

 

[wifi]

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)

$(\nabla \times \vec{v}) \cdot d\vec{a} = [6(x-y)^2 \hat{z}] \cdot [dx \ dy \ \hat{z}] = 6(x-y)^2dx \ dy$

$\int (\nabla \times \vec{v}) \cdot d\vec{a} = \int \int 6(x-y)^2dx \ dy = 6 \int \int (x-(2x-2))^2dx \ dy = 6 \int \int (2-x)^2dx \ dy = 6 \int \int (x^2-4x+4) dx \ dy$.

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

 

[wifi]

So with the limits, it should be this $6 \int_0^{-2} \int_0^{\frac{2+y}{2}} (x^2-4x+4) dx \ dy$, correct?

 

[CAF123]

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.

 

[wifi]

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

 

[wifi]

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

 

[CAF123]

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.

 

[wifi]

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?

 

[CAF123]

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?

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.

 

[wifi]

CAF123, thanks so much! I ended up getting 7 for the answer for both the surface integral & line integral. I handed in my work already, but I'll try to post my solution on here.