Stokes' Theorem for F(x,y,z) over S: Solving for dA on a Bounded Surface

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In summary: For a surface in two dimensions, there is only one normal, but for a surface in one dimension, there are multiple normal vectors. Thanks for the clarification!
  • #1
Ineedahero
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1

Homework Statement


Verify Stokes' Theorem

F(x,y,z) = (xz,-y,x2y) and S is the surface bounded by the planes x = 0, y = 0, z = 0, and 2x + y + 2z = 8, excluding the part contained in the xz-plane.


Homework Equations


Stokes' Theorem:
∫∫A∇xF dA = ∫dAF dR


The Attempt at a Solution


Did the right hand side, got my answer of 4.

The left hand side is where I have trouble.
Curl F = {x2, x-2xy, 0}

How do I define dA? It's three separate planes. ...?
What are my bounds?

Thank you for your help
 
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  • #2
Ineedahero said:

Homework Statement


Verify Stokes' Theorem

F(x,y,z) = (xz,-y,x2y) and S is the surface bounded by the planes x = 0, y = 0, z = 0, and 2x + y + 2z = 8, excluding the part contained in the xz-plane.


Homework Equations


Stokes' Theorem:
∫∫A∇xF dA = ∫dAF dR


The Attempt at a Solution


Did the right hand side, got my answer of 4.

The left hand side is where I have trouble.
Curl F = {x2, x-2xy, 0}

How do I define dA? It's three separate planes. ...?
What are my bounds?

Thank you for your help

You do three separate integrals. In the xy plane dA = dydx, in the yz plane it is dydz and you integrate over the corresponding triangles. Do you know how to calculate dA for the slanted plane, like you would do for any other surface?
 
  • #3
Thanks for the help!

Okay...here's where I got:

You said I need three integrals. Here they are:

1.) xy plane, z = 0
∫∫(x2, x-2xy, 0) dot (0,0,1)dxdy, where (0,0,1) is perpendicular vector to the plane
= 0

2.) yz plane, x = 0
0408-2z (x2, x-2xy, 0) dot (1,0,0)dydz, where (1,0,0) is perpendicular vector. Are these bounds correct?
x = 0, so the integral is zero. But were the bounds correct?

3.) plane 2x + y + 2z = 8
0408-2x (x2, x-2xy, 0) dot (2,1,2) dydx, where (2,1,2) is perpendicular vector to the plane. I know this is wrong because it does not account for z. How should I do it?
 
  • #4
Actually, the boundary consists of four separate planes: x= 0, y= 0, z= 0, and 2x + y + 2z = 8.

On x= 0, the yz-plane, the outward pointing normal is (-1, 0, 0) and dS= dydz.

On y= 0, the xz-plane, the outward pointing normal is (0, -1, 0) and dS= dxdz.

On z= 0, the xy-plane, the outward pointing normal is (0, 0, -1) and dS= dxdy.

On 2x + y + 2z = 8 we can write y= 8- 2x- 2z so the "position vector" is [tex]\vec{r}= x\vec{i}+ (8- 2x- 2z)\vec{j}+ z\vec{k}[/tex]. [tex]\vec{r}_x= \vec{i}- 2\vec{j}[/tex] and [tex]\vec{r}_z= -2\vec{j}+ \vec{k}[/tex]. The cross product and so the outward pointing normal is [tex]2\vec{i}+\vec{j}+ 2\vec{k}[/tex] and dS= dxdz.
 
  • #5
Ineedahero said:
Thanks for the help!

Okay...here's where I got:

You said I need three integrals. Here they are:

1.) xy plane, z = 0
∫∫(x2, x-2xy, 0) dot (0,0,1)dxdy, where (0,0,1) is perpendicular vector to the plane
= 0

2.) yz plane, x = 0
0408-2z (x2, x-2xy, 0) dot (1,0,0)dydz, where (1,0,0) is perpendicular vector. Are these bounds correct?
x = 0, so the integral is zero. But were the bounds correct?

I have a nit-pick on both 1 and 2. Actually it's more than a nit-pick; it is important. Stokes theorem refers to an oriented surface. You have not given the orientation of the surface. If you assume it is oriented outward from the (almost) enclosed volume, your normals are both incorrect. Do you see why? Luckily, the answer is 0 anyway, but that's just luck. And yes, the limits on 2 are correct but you didn't show the limits on 1.

3.) plane 2x + y + 2z = 8
0408-2x (x2, x-2xy, 0) dot (2,1,2) dydx, where (2,1,2) is perpendicular vector to the plane. I know this is wrong because it does not account for z. How should I do it?

You use the formula for dS for a surface. For z = f(x,y), what is that formula?
 
  • #6
HallsofIvy said:
Actually, the boundary consists of four separate planes: x= 0, y= 0, z= 0, and 2x + y + 2z = 8.
...

The cross product and so the outward pointing normal is ## 2\vec{i}+\vec{j}+ 2\vec{k}##
and dS= dxdz.

No, it is an open surface. And dS isn't dxdz on the plane.
 
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  • #7
LCKurtz said:
I have a nit-pick on both 1 and 2. Actually it's more than a nit-pick; it is important. Stokes theorem refers to an oriented surface. You have not given the orientation of the surface. If you assume it is oriented outward from the (almost) enclosed volume, your normals are both incorrect. Do you see why? Luckily, the answer is 0 anyway, but that's just luck. And yes, the limits on 2 are correct but you didn't show the limits on 1.



You use the formula for dS for a surface. For z = f(x,y), what is that formula?

Yes, I see the error - my normal vectors are indeed perpendicular, but they point inward. Thanks for noticing that!

And as for the third integral...

z = (1/2)(8-2x-y)...when z = 0, ---> 2x + y = 8
Thus, 0 <= y <= 8 - 2x
And when y= 0 and z = 0 ---> x = 4
Hence, 0 <= x <= 4

Does that imply that my original integral was indeed correct?
 
  • #8
Ineedahero said:
Yes, I see the error - my normal vectors are indeed perpendicular, but they point inward. Thanks for noticing that!

And as for the third integral...

z = (1/2)(8-2x-y)...when z = 0, ---> 2x + y = 8
Thus, 0 <= y <= 8 - 2x
And when y= 0 and z = 0 ---> x = 4
Hence, 0 <= x <= 4

Does that imply that my original integral was indeed correct?

dS does not equal dxdy when you are on the slanted plane.
 
  • #9
So would it be dxdydz??

I am wary to do that because I am under the impression that dxdydz does not represent 2-dimensional surfaces that are planes.

But...I presume then that the integral would be a TRIPLE integral, with the bounds of z: 0 <= z <= (1/2)(8-2x-y), and the other bounds for x and y would remain as I stated.

I did have a strong suspicious that setting z = 0 was erroneous...
 
  • #10
You can solve the plane for z and get the form z = f(x,y). Look in your book to find the formula for dS when z = f(x,y).
 
  • #11
I tried looking in my book but I wasn't really sure where to look...I looked at a couple example problems of Stokes' Theorem, but none of them involved planes...

If dS is not dxdy when z = f(x,y) and it's not dxdydz...then what could it possibly be??
 
  • #13
In his example, f is a scalar function. In my example, it is a vector function.

From that, I presume that I use the perpendicular vector and not its magnitude. I noticed that after he multiplies by the magnitude of the perpendicular vector, he made dA = dy dz, and he had x = g(y,z).

This all leads me to think that this is the correct integral:

0408-2x( (x/2)(8-2x-y) , -y , x2y ) dot (2,1,2) dy dx

But you said that dy dx is not correct??
 
  • #14
Ineedahero said:
In his example, f is a scalar function. In my example, it is a vector function.

From that, I presume that I use the perpendicular vector and not its magnitude. I noticed that after he multiplies by the magnitude of the perpendicular vector, he made dA = dy dz, and he had x = g(y,z).

This all leads me to think that this is the correct integral:

0408-2x( (x/2)(8-2x-y) , -y , x2y ) dot (2,1,2) dy dx

But you said that dy dx is not correct??

Well that integrand certainly isn't correct because it is supposed to be ##\nabla \times \vec F## in there. Perhaps you are getting confused between ##dA##, ##dS## and ##d\vec S##. Stokes theorem says$$
\iint_S\nabla \times \vec F\cdot d\vec S =\int_C \vec F\cdot d\vec R$$where ##\vec R## is a parameterization of the bounding curve.

Let's parameterize the plane as ##x=x,\, y = y,\, z = \frac{8-2x-y} 2## so$$
\vec r(x,y) = \langle x, y, \frac {8-2x-y} 2\rangle$$In this setting ##dA = dydx##.
The scalar element of surface area is defined as ##dS = |\vec r_x\times\vec r_y|dydx##. The vector area element is defined as$$
d\vec S =\hat n dS =\frac{\pm\vec r_x\times\vec r_y}{|\vec r_x\times\vec r_y|}
|\vec r_x\times\vec r_y|dydx =\pm\vec r_x\times\vec r_ydydx$$where ##\hat n## is a unit normal to the surface in the proper direction given by the orientation (choose the ##\pm## accordingly).

So calculate ##d\vec S = \pm\vec r_x\times\vec r_ydydx## with the correct sign, dot ##\nabla\times \vec F## into it and use that as your integrand. By the way I think you should be getting 32/3 on both sides for your answer.
 
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  • #15
Of course - it should be the curl, I suppose I was a little exhausted when I did that.

And yes, I've actually reworked the right hand side, and got a result of 32/3.

And, rather than compute the cross product, could I not deduce the normal vector from the equation of the plane? 2x + 1y + 2z = 8 ----> (2,1,2)/3 -----> positive for outward pointing normal.

EDIT: And divide by 3 to make it a unit vector
 
  • #16
Ineedahero said:
Of course - it should be the curl, I suppose I was a little exhausted when I did that.

And yes, I've actually reworked the right hand side, and got a result of 32/3.

And, rather than compute the cross product, could I not deduce the normal vector from the equation of the plane? 2x + 1y + 2z = 8 ----> (2,1,2)/3 -----> positive for outward pointing normal.

EDIT: And divide by 3 to make it a unit vector

Yes. But then you must remember that ##dS \ne dydx##. It is ##|r_x\times \vec r_y| dydx## so you still need the partials.
 
  • #17
Ahhhhhhhhhhhh, finally, I got the answers to match.

Thank you so much for your help! You really cleared everything up. Thanks!
 
  • #18
I totally confused with this question. can explain me with the sketched diagram please?
 
  • #19
Indhu Rani said:
I totally confused with this question. can explain me with the sketched diagram please?
This is a two year old thread. Please start a new thread with your question and use the homework template.
 

Related to Stokes' Theorem for F(x,y,z) over S: Solving for dA on a Bounded Surface

1. What is Stokes' Theorem for F(x,y,z) over S?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed, bounded surface to the line integral of the same vector field over the boundary curve of the surface. It is named after Irish mathematician George Stokes.

2. How is Stokes' Theorem used to solve for dA?

Stokes' Theorem states that the surface integral of a vector field F over a closed, bounded surface S is equal to the line integral of the same vector field over the boundary curve of S. This allows us to calculate the surface integral by instead calculating the line integral, which is often easier to solve for. Therefore, Stokes' Theorem is used to solve for dA by converting the problem from a surface integral to a line integral.

3. What is a bounded surface in the context of Stokes' Theorem?

A bounded surface is a surface that has a well-defined boundary curve, meaning that the surface does not extend to infinity and has a finite area. In the context of Stokes' Theorem, the surface must be closed and bounded in order for the theorem to hold true.

4. Are there any limitations to using Stokes' Theorem for F(x,y,z) over S?

Stokes' Theorem is only applicable to closed, bounded surfaces, as mentioned earlier. It also requires that the vector field F is continuously differentiable on the surface S and its boundary curve. Additionally, the surface S must be oriented consistently with the boundary curve in order for Stokes' Theorem to be valid.

5. How is Stokes' Theorem related to other fundamental theorems in vector calculus?

Stokes' Theorem is closely related to two other fundamental theorems in vector calculus: the Divergence Theorem and Green's Theorem. All three theorems relate a surface integral to a line integral, but they differ in the dimensionality of the space and the type of integral used. Stokes' Theorem is a generalization of Green's Theorem and can also be seen as a higher-dimensional version of the Fundamental Theorem of Calculus.

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