F(x, y) represents a velocity field of a fluid over a surface S defined by z = 6 − 3x − 2y. If the magnitude of the velocity in the direction of the unit normal vector, n̂, on S is 3z⁄√14, compute the flux of F(x, y) over the surface S in the first octant oriented upward, using the projection of S on the xy - plane

Question

F(x, y) represents a velocity field of a fluid over a surface S defined by z = 6 − 3x − 2y. If the magnitude of the velocity in the direction of the unit normal vector, n̂, on S is 3z⁄√14, compute the flux of F(x, y) over the surface S in the first octant oriented upward, using the projection of S on the xy - plane

Answer 1

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx

boundary in first quadrant means x from 0 to 2 and y from 0 to 3
(sketch that thing)

Answer 2

n = 3 i +2 j + k

we want F dot n
= |F| |n| cos angle
but we are given that |F| cos angle is 3 z/sqrt14
so what is |n|?
sqrt (9+4+1) = sqrt 14
ha, that helps :)

so F dot n = 3 z
so
do the integral of
3 z
over the surface

Answer 3

thanks for your help. do you mean F.n=3z/sqrt14??

Answer 4

Thanks sir Damon.. Understood with ur explanation..

Answer 5

No, because multiply by sqrt 14 which is |n|

Answer 6

Advanced Engineering Mathematics

By Dennis G. Zill, Michael R. Cullen

you can get it from preview mode on google books page 529 and 530 it got the similar question... be sure to check it out. The answer is 18

Answer 7

To compute the flux of the vector field F(x, y) over the surface S, we can use the surface integral:

Flux = ∬(F(x, y) ⋅ n̂) dS

where n̂ is the unit normal vector to the surface S, dS is an infinitesimal area element on S, ⋅ denotes the dot product, and ∬ denotes the double integral over the surface.

In this case, the surface S is defined by the equation z = 6 - 3x - 2y. To find the unit normal vector n̂, we can first find the gradient vector ∇z of the surface, and then normalize it.

1. Finding the gradient vector ∇z:
∇z = (dz/dx, dz/dy, dz/dz) = (-3, -2, 1)

2. Normalizing the gradient vector to obtain the unit normal vector n̂:
n̂ = ∇z / ||∇z|| = (-3, -2, 1) / √(9 + 4 + 1) = (-3/√14, -2/√14, 1/√14)

Now that we have the unit normal vector n̂, we can compute the flux of the vector field F(x, y) over the surface S using a double integral.

3. Projecting S onto the xy-plane:
To compute the flux of F(x, y) over S, we need to determine the projection of S onto the xy-plane. The equation of the surface S can be rewritten as z = 6 - 3x - 2y. Setting z = 0, we get:
0 = 6 - 3x - 2y
3x + 2y = 6

This represents the projection of S onto the xy-plane.

4. Computing the flux using a double integral:
The flux of F(x, y) over the surface S is given by the double integral:

Flux = ∬(F(x, y) ⋅ n̂) dS

Since the surface S is oriented upward in the first octant, we only consider the region of the projection that lies in the first octant.

∬(F(x, y) ⋅ n̂) dS = ∬(F(x, y) ⋅ n̂) dA

where dA is an infinitesimal area element in the xy-plane.

Now, we need to parametrize the region of the projection that lies in the first octant. We can use u and v as the variables of integration, where u represents x and v represents y.

- For the limits of u, we have: 0 ≤ u ≤ 2 (from the equation 3x + 2y = 6).
- For the limits of v, we have: 0 ≤ v ≤ (6 - 3u)/2 (from the equation 3x + 2y = 6).

Substituting the parametrization u = x and v = y into the vector field F(x, y), we get:

F(u, v) = <F1(u, v), F2(u, v), F3(u, v)>

Now, we can compute the dot product between F(u, v) and n̂, and integrate over the region of projection:

Flux = ∫[0 to 2]∫[0 to (6 - 3u)/2] (F(u, v) ⋅ n̂) dv du

Evaluate this double integral to find the flux of F(x, y) over the surface S in the first octant oriented upward, using the projection of S on the xy-plane.