# Find the field flow using integrals

1. Jun 14, 2010

### kliker

1. The problem statement, all variables and given/known data
find the field's F flow where F = (y,-z,2) through the surface S where S is 4x+2y+z=6 and x,y,z>=0

3. The attempt at a solution

we know that

[PLAIN]http://img195.imageshack.us/img195/4971/asdqg.gif [Broken]

we know F

N is the derivative of 4x+2y+z-6 = g(x)

actually it's N = (dg/dx,dg/dy,dg/dz)

now what i want to ask is how to find the limits of the integral i mean from what point to what point i should integrate

what I can think of is because x,y,z are bigger than 0 hence they are possitive

we will have the same points for dy and dx which will be from 0 to +oo

so can i just say

[PLAIN]http://img514.imageshack.us/img514/351/asdd.gif [Broken]

find the integral and then take the limit using xn -> oo and yn -> oo?

edit: but i have a z, what am i going to do with z?

Last edited by a moderator: May 4, 2017
2. Jun 14, 2010

You need to evaluate a surface integral. Rewrite the equation of the plane in the form z = f(x, y), plug it into your integral, and find the triangular region over which you need to integrate, together with the appropriate boundaries of integration.

3. Jun 14, 2010

By the way, applying the divergence theorem yields the answer very quickly.

4. Jun 14, 2010

### HallsofIvy

4x+ 2y+ z= 6 is a plane. Since "x, y, z>= 0", that is, all coordinates must be non-negative, you want the portion of the plane that is in the first octant. It's boundary consists of points where some of x, y, z are equal to 0. Specifically in the xy-plane, where z= 0, the triangular region radou referred to has boundaries x= 0, y= 0, and 4x+ 2y= 6 so that y= 2x- 3. y will also be 0 when 2x- 3=0 or x= 3/2.

x goes from 0 to 3/2 and, for every x, y goes from 0 to 2x- 3.

5. Jun 14, 2010

### kliker

one last question

should I replace z = 6-2y-4x if i have a z in the integral?

6. Jun 14, 2010

### LCKurtz

No, it doesn't. You don't have an enclosed volume.

Yes. But I have a couple of questions for you. First, did you check the orientation of the surface and do you know you are calculating the flow in the proper direction?

Second, you apparently got your normal vector by taking the gradient of the plane equation, which gives the coefficients. But [flux] surface integrals are of the form

$$\int\int_R \vec F \cdot \hat n\ dS$$

using a unit normal, which your N isn't. Now it turns out that your integrand is correct but my question to you is do you know why or did you just get lucky? Why did you not use the unit normal? Are you using something you didn't tell us?

Last edited: Jun 14, 2010
7. Jun 14, 2010

### HallsofIvy

But Stoke's theorem could convert it into an integration around the bounding triangle. That probably would be harder, not simpler!

8. Jun 15, 2010

### kliker

We ve learned in Math that field flow has this kind of integral, that's why I used it

the reason i used simple F and N is because i dont know how to create a vector F and N on my keyboard

9. Jun 15, 2010

### LCKurtz

I wasn't asking because of the typing. I was asking because you didn't use a unit vector in your calculation. Your N is not a unit vector. And dS is not dxdy. I was wondering if you could explain about that. And you didn't answer the question about the orientation of the surface. Which way is it oriented and does your choice of normal agree with it?