Vector Surface Integrals: Limits and Unit Vectors

In summary, the conversation is about calculating the flux of a vector field through a square area element centered on the x-axis. The equation for the vector surface integral is provided, but there is uncertainty about determining the upper and lower limits for the z and y paths. The unit vector, n, is also mentioned, and it is equal to the unit vector x which is equal to the cross product of unit vectors y and z. The given information includes the vector field V = xyz, the length of one side of the square being 1, and the center of the square being at x = x0. The question of what the limits should be for y and z is discussed, with the possibilities being (-1/2, 1/
  • #1
Void123
141
0

Homework Statement



I am given a vector field associated with a square area of a certain side, let's call this side dx, centered on the x-axis at a certain point, say x = x1. The sides of this cross sectional square are parallel to the axis of y and z. I have to compute the flux of this vector field, or rather, the vector surface integral.

Now I understand how to set up the equation, substitute the relevant variables, and such. What I'm not totally certain of is how to determine the upper and lower limits over the z and y paths.

Also, what does the unit vector, n, do?

Homework Equations



V n dS, where dS = dydz, and V and n are vectors.

The Attempt at a Solution



As I said, I have already set up the (double) integral according to the above formula, but I need to determine the limits of the definite integral.
 
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  • #2
I think they are usually provided. Could you post one question where you think they are not provided?
 
  • #3
rootX said:
I think they are usually provided. Could you post one question where you think they are not provided?

I am given a vector field that permeates a square area element of a side (of certain value), centered on the x-axis (at a given point on x). The square's sides are parallel to the y and z axes. The unit vector n is equal to the unit vector x, which is equal to the cross product of unit vectors y and z.

That is all I have and I am told to calculate the vector surface integral.

I cannot think of a way to determine the limits. But other than that, the expression is pretty straightforward.
 
  • #4
Void123 said:
I am given a vector field that permeates a square area element of a side (of certain value), centered on the x-axis (at a given point on x). The square's sides are parallel to the y and z axes. The unit vector n is equal to the unit vector x, which is equal to the cross product of unit vectors y and z.

That is all I have and I am told to calculate the vector surface integral.

I cannot think of a way to determine the limits. But other than that, the expression is pretty straightforward.

Please "post one question where you think they are not provide". Don't just repeat what you said before, post the actual question!
 
  • #5
HallsofIvy said:
Please "post one question where you think they are not provide". Don't just repeat what you said before, post the actual question!

That is the actual question. Only thing I can add is that the square area has a side of 1, centered on the x-axis at x = 1 (don't know how much this helps). It looks like a flux problem.
 
  • #6
Void123 said:
I am given a vector field that permeates a square area element of a side (of certain value), centered on the x-axis (at a given point on x). The square's sides are parallel to the y and z axes. The unit vector n is equal to the unit vector x, which is equal to the cross product of unit vectors y and z.

That is all I have and I am told to calculate the vector surface integral.

I cannot think of a way to determine the limits. But other than that, the expression is pretty straightforward.

If you are "given" a square, you must have been given the length of one of its sides, call it w. If you know at it as some point on the x-axis, say x = a, then you know the square must have the parametric form < a, x, y > where x and y vary over the square. Now you just have to use what you are given by the placement of the square. Does x go from 0 to w or from -w/2 to w/2 or what? Ditto for y. Draw a picture.

You haven't given us the vector field. If the vector field is constant it might not matter where the square is located, but usually it would matter.
 
  • #7
The vector field is V = xyz (in the xth direction).

If the side of the square is 1 and its centered around the x-axis with the sides parallel to the y and z axes, then wouldn't the limits be (0, 1/2) for y and z, respectively?
 
Last edited:
  • #8
Void123 said:
The vector field is V = xyz (in the xth direction).

If the side of the square is 1 and its centered around the x-axis with the sides parallel to the y and z axes, then wouldn't the limits be (0, 1/2) for y and z, respectively?

I collected that
1) <1,0,0> is the normal
2) surface area is a square with length of 1 and center of x0
3) the vector field is V = xyz

Draw a square with center at x0, so are the limits for y and z from -1/2 to 1/2 or 0 to 1/2?

P.S. you should provide all the information in one post. Please note that other people cannot read your mind.
 
  • #9
rootX said:
I collected that
1) <1,0,0> is the normal
2) surface area is a square with length of 1 and center of x0
3) the vector field is V = xyz

Draw a square with center at x0, so are the limits for y and z from -1/2 to 1/2 or 0 to 1/2?

P.S. you should provide all the information in one post. Please note that other people cannot read your mind.

I am not sure about the limits. I am trying to think of any information that would be indicative of what the limits should be.

Also, I apologize if I have not been clear.
 
  • #10
rootX said:
I collected that
1) <1,0,0> is the normal
2) surface area is a square with length of 1 and center of x0
3) the vector field is V = xyz

Draw a square with center at x0, so are the limits for y and z from -1/2 to 1/2 or 0 to 1/2?

P.S. you should provide all the information in one post. Please note that other people cannot read your mind.

With regard to 1) ..so dS = <1, 0, 0> dy dz, right?

With regard to 2) The [bold]surface[\bold] is a square. Its area is a number. And when you say "center at x0" do you mean (x0, 0, 0)?

With regard to 3) V = xyz is not a vector field; it is an ordinary scalar function.
I guess you mean V = <xyz, 0 , 0 >

As for whether the limits on y and z are -1/2 to 1/2 or 0 to 1/2, which choice would describe a square of side 1 centered at (x0, 0, 0)?

In my reply I said:

Kurtz said:
If you are "given" a square, you must have been given the length of one of its sides, call it w. If you know at it as some point on the x-axis, say x = a, then you know the square must have the parametric form < a, x, y > where x and y vary over the square. Now you just have to use what you are given by the placement of the square. Does x go from 0 to w or from -w/2 to w/2 or what? Ditto for y. Draw a picture.[/qoute]

I have a typo there; the parametric from of the square would be <a, y, z> where y and z vary over the range discussed above.

Do you have enough to solve it now?
 
  • #11
Everything you have stated sounds legitimate to me. The only confusion I have is deciding between (0,w) or (-w/2, w/2). I am sure I am missing something very trivial here which ought to resolve this.
 
  • #12
Void123 said:
Everything you have stated sounds legitimate to me. The only confusion I have is deciding between (0,w) or (-w/2, w/2). I am sure I am missing something very trivial here which ought to resolve this.

If your square is centered at (x0, 0, 0) and is parallel to the yz plane, as far as the yz plane is concerned, it is centered at (0,0). Draw a picture of a square centered at (0,0) whose sides have length w. That should answer your question.
 
  • #13
LCKurtz said:
If your square is centered at (x0, 0, 0) and is parallel to the yz plane, as far as the yz plane is concerned, it is centered at (0,0). Draw a picture of a square centered at (0,0) whose sides have length w. That should answer your question.

Thank you!
 

1. What is a vector surface integral?

A vector surface integral is a type of mathematical calculation used to determine the total flux (flow) of a vector field across a given surface. It takes into account both the magnitude and direction of the vector at each point on the surface.

2. How is a vector surface integral calculated?

To calculate a vector surface integral, the surface is divided into small elements and the flux across each element is calculated. The sum of all these flux values gives the total flux across the entire surface. This can be represented mathematically using a double integral.

3. What are some real-world applications of vector surface integrals?

Vector surface integrals are used in many fields such as physics, engineering, and fluid dynamics. They can be used to calculate the flow of electric fields, magnetic fields, and fluids across a given surface. They are also important in studying the behavior of electromagnetic waves and in designing structures that can withstand fluid flow.

4. What is the difference between a scalar surface integral and a vector surface integral?

A scalar surface integral calculates the total value of a scalar function over a given surface, while a vector surface integral calculates the total flux of a vector field across a surface. In other words, a scalar surface integral results in a single value, whereas a vector surface integral results in a vector quantity.

5. Are there any common mistakes to avoid when working with vector surface integrals?

Some common mistakes to avoid when working with vector surface integrals include not properly defining the surface or the vector field, forgetting to include the direction of the vector in the calculation, and making incorrect assumptions about the symmetry of the surface or the vector field. It is important to carefully set up the integral and double-check all calculations to ensure accuracy.

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