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
To compute the flux of the velocity field F(x, y) over the surface S in the first octant oriented upward, we need to perform the following steps:
Step 1: Find the unit normal vector n̂ to the surface S:
The surface S is defined by the equation z = 6 − 3x − 2y. To find the unit normal vector n̂, we need to take the partial derivatives of S with respect to x and y and find their cross product.
Taking the partial derivatives, we have:
∂S/∂x = -3
∂S/∂y = -2
The partial derivatives represent the components of the tangent vector to the surface S. By taking the cross product, we can find the normal vector n̂:
n̂ = ∂S/∂x × ∂S/∂y
= (-3i + 0j + k) × (0i - 2j + k)
= (-2k + 6j + 0i)
Since we want n̂ to be a unit vector, we need to normalize it by dividing each component by its magnitude. The magnitude of n̂ can be found using the formula ||n̂|| = sqrt(a^2 + b^2 + c^2) for a vector (a, b, c).
||n̂|| = sqrt((-2)^2 + 6^2 + 0^2)
= sqrt(4 + 36)
= sqrt(40)
= 2√10
Thus, the unit normal vector n̂ is:
n̂ = (-2/2√10)k + (6/2√10)j + 0i
= -(1/√10)k + (3/√10)j
Step 2: Project the surface S onto the xy-plane:
To project the surface S onto the xy-plane, we can simply ignore the z-coordinate of the surface equation. In this case, we have z = 6 − 3x − 2y.
Step 3: Compute the flux of F(x, y) over the projected surface:
The flux of F(x, y) over the surface S can be computed using the flux integral formula:
Flux = ∬_S F(x, y) · n̂ dS
In this case, we are given that the magnitude of the velocity in the direction of the unit normal vector n̂ on S is 3z/√14, so we can write F(x, y) as:
F(x, y) = (3z/√14) n̂
The magnitude of F(x, y) is therefore 3z/√14. Since we have projected the surface S onto the xy-plane, the area element dS becomes dA.
Substituting these values into the flux integral formula, we have:
Flux = ∬_A (3z/√14) n̂ · dA
Step 4: Evaluate the flux integral over the projected surface A:
To evaluate the flux integral, we need to find the limits of integration for the projected surface A.
In the first octant, x and y are both positive, so the limits of integration for x and y will be determined by the bounds of the region in the xy-plane where the projection occurs.
From the equation z = 6 − 3x − 2y, we can solve for z in terms of x and y:
z = 6 − 3x − 2y
Since we have projected the surface onto the xy-plane, z = 0. Substituting this into the equation, we have:
0 = 6 − 3x − 2y
Simplifying, we get:
3x + 2y = 6
This represents a straight line in the xy-plane. To find the limits of integration, we need to determine the bounds where this line intersects the positive x and y axes.
Setting x = 0, we have:
2y = 6
y = 3
Setting y = 0, we have:
3x = 6
x = 2
Therefore, the limits of integration for x are from 0 to 2, and for y they are from 0 to 3.
Finally, we can evaluate the flux integral using these limits of integration.