Flux of the vector field

In summary, the problem involves finding the flux of the vector field F=7xi+yj+zk across the plane z+4x+2y=12. The solution involves taking the dot product of F with the unit normal to the plane, and integrating over the projected region in the xy-plane. The ends of integration can be found by projecting the plane onto the xy-plane and finding the limits of integration.
  • #1
UrbanXrisis
1,196
1
Let S be the part of the plane 3x+y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+2j+3k across the surface S.

[tex]\int \int F\cdot dS = \int int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex]

[tex]\int \int \left( -4(-3)-2(-1)+4-3x-y \right) dA[/tex]
[tex]\int \int \left( 18-3x-y \right) dA[/tex]

how do I find the ends of integration?
 
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  • #2
UrbanXrisis, I really don't understand what you did.

What I would have done is
[tex]\int \int F\cdot dS = \int \int \left(F\cdot \hat{n}\right) \left(\frac{dxdy}{\hat{n}\cdot\hat{k}}\right)[/tex]

where, [itex] \hat{n} [/itex] is the unit normal to the plane. That is, to evaluate the surface integral, I'm simply projecting the surface to the x-y plane and integrating.

Now to find the ends of integration, just project the surface to the x-y plane. In this case, you will get a triangle. From that, you can find the limits of integration
 
  • #3
UrbanXrisis said:
Let S be the part of the plane 3x+y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+2j+3k across the surface S.

[tex]\int \int F\cdot dS = \int int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex][\quote]
? There were no "P", "Q", "R" in the problem! If you MEAN 3x+ y+ z= 4 so P= 3, y= 1, z= 1 then say that!

[tex]\int \int \left( -4(-3)-2(-1)+4-3x-y \right) dA[/tex]
[tex]\int \int \left( 18-3x-y \right) dA[/tex]
Again mystifying! Assuming you are doing the integration by projecting down into the xy- plane, then [itex]d\sigma^{/rightarrow}= (3i+ j+ k)dxdy so the integrand is (4i+ 2j+ 3k) . (3i+ j+ k) dxdy= 17dxdy

how do I find the ends of integration?
The limits of integration? Projecting the plane, 3x+y+z=4, into the xy-plane gives 3x+ y+ 0= 4 or y= 4- 3x. When y= 0, x= 4/3 so you would integrate with respect to y from 0 to 4- 3x and then with respect to x from 0 to 4/3.

However, the integral of '17 dA' is just 17A, or 17 times the area of that triangle. What is the area of a triangle with base 4/3 and height 4?
 
  • #4
I have a similar question to this, with F = 7xi+yj+zk and plane z + 4x + 2y = 12. so what I did was (7xi+ yj+ zk) . (4i+ 2j+ k), is this correct?
 
  • #5
anyone?
 

1. What is a flux of a vector field?

Flux of a vector field is a measure of the flow of a vector field through a surface. It represents the quantity of the vector field passing through a given surface in a unit time.

2. How is the flux of a vector field calculated?

The flux of a vector field is calculated by taking the dot product of the vector field and the unit normal vector to the surface, and integrating this product over the surface.

3. What is the physical significance of flux of a vector field?

The flux of a vector field is an important concept in fluid dynamics and electromagnetics. It helps in understanding the flow of fluids and the distribution of electric or magnetic fields through a surface.

4. Can the flux of a vector field be negative?

Yes, the flux of a vector field can be negative. This occurs when the vector field is flowing in the opposite direction to the surface normal vector, indicating that the flow is going inwards rather than outwards.

5. How does the divergence theorem relate to flux of a vector field?

The divergence theorem states that the flux of a vector field through a closed surface is equal to the triple integral of the divergence of the vector field over the region enclosed by the surface. This provides a useful way to calculate the flux of a vector field in certain situations.

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