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

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In summary, we are using Stokes' Theorem to check the vector function ## v=(x-y)^3 \hat{x} + (x-y)^3 \hat{y} ## by computing both a surface integral and a line integral. We split the path integral into three parts and obtained a different answer than the previous calculation. However, this is acceptable as the integral is a measure of the net flux through a surface. The limits of integration for x and y were determined in order to follow the counter-clockwise path. The final answer for both integrals was 7.
  • #1
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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?
 

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  • #2
wifi said:
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
Thanks Dick. I computed the surface integral & I got 101/6. Is this correct?
 
  • #4
wifi said:
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
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.
 
  • #6
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 ##.
 
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  • #7
Anyone?
 
  • #8
wifi said:
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
I skipped the other 2 integrals because I believe they come out to be 0.
 
  • #10
## \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 ##
 
  • #11
wifi said:
I skipped the other 2 integrals because I believe they come out to be 0.

Can you show your work?
 
  • #12
Certainly! I'll type it out right now. Thanks for the help by the way!
 
  • #13
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?
 
  • #14
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?
 
  • #15
wifi said:
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
So that means I must integrate over y first and then x?
 
  • #17
The limits would then be y: 0 -> 2x-2 and x: 1 -> 0?
 
  • #18
wifi said:
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
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?
 
  • #20
wifi said:
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.
 
  • #21
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.
 

What is Stokes's Theorem for Vector Function v=(x-y)^3?

Stokes's Theorem for Vector Function v=(x-y)^3 is a mathematical theorem that relates the surface integral of a vector function over a closed surface to the line integral of the same function over the boundary curve of the surface. It is a fundamental theorem in vector calculus and is often used in physics and engineering applications.

What is the significance of using v=(x-y)^3 in Stokes's Theorem?

The vector function v=(x-y)^3 is often used in Stokes's Theorem because it represents a conservative vector field, meaning the line integral of the function over any closed curve is equal to zero. This simplifies the calculations and makes the theorem more applicable in real-world problems.

How is the area calculation related to Stokes's Theorem for Vector Function v=(x-y)^3?

Stokes's Theorem can be used to calculate the area of a surface by taking the surface integral of the function v=(x-y)^3 over the surface. This is based on the concept that the flux of the vector field through the surface is equal to the line integral of the function over the boundary curve of the surface, which in this case represents the perimeter of the surface.

Can Stokes's Theorem for Vector Function v=(x-y)^3 be used to calculate the perimeter of a surface?

Yes, Stokes's Theorem can be used to calculate the perimeter of a surface by taking the line integral of the function v=(x-y)^3 over the boundary curve of the surface. This is based on the concept that the line integral of a conservative vector field over a closed curve is equal to the area enclosed by the curve.

What are some practical applications of Stokes's Theorem for Vector Function v=(x-y)^3?

Stokes's Theorem has numerous practical applications in various fields such as fluid dynamics, electromagnetism, and engineering. It is often used to calculate the circulation of a fluid around a closed path, the work done by a magnetic field on a moving charge, and the force exerted by a fluid on a solid object. It is also used in the study of fluid flow and heat transfer in pipes and channels.

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