How Do You Calculate the Upward Flux of a Vector Field Over a Plane Section?

[hils0005]

Homework Statement

Let S be part of the plane 3x + 2y + z =9 lying in the first octant. Calculate the upward flux of vectorF(x,y,z) = x i + xy j + xz k

Homework Equations

z=9-3x-2y
dz/dx=-3
dz/dy=-2

F(dot)(-dz/dx i - dz/dy j + k)

The Attempt at a Solution

x i + xy j + xz k (dot) (3i +2j + k)=3x + 2xy + xz
I then substituted z: =3x + 2xy + 9x -3x^2 - 2xy

$$\int$$$$\int$$ -3x^2 + 12x dydx

0 $$\leq$$ y $$\leq$$ -4.5x/3 + 4.5
0 $$\leq$$ x $$\leq$$ 3

Any help on whether this is correct or not would be apprecitated

 

[mighty2000]

.

Your attempt at a solution is on the right track, but there are a few errors and missing steps. Here is a step-by-step solution:

1. Write the given equation in terms of z: z = 9 - 3x - 2y.
2. Calculate the partial derivatives of z with respect to x and y: dz/dx = -3 and dz/dy = -2.
3. Use the given vector field F to calculate the dot product with the normal vector of the plane: F · (-3i - 2j + k) = -3x - 2xy + xz.
4. Substitute the expression for z from step 1 into the dot product: -3x - 2xy + x(9 - 3x - 2y) = -3x - 2xy + 9x - 3x^2 - 2xy.
5. Simplify the expression: -3x^2 + 12x - 4xy.
6. Write the integral for the upward flux over the region in the first octant: ∫∫ (-3x^2 + 12x - 4xy) dydx, with the limits of integration as 0 ≤ y ≤ (-4.5x/3) + 4.5 and 0 ≤ x ≤ 3.
7. Integrate with respect to y first: ∫ (-3x^2y + 12xy - 2xy^2) dy, with the limits of integration 0 and (-4.5x/3) + 4.5.
8. Substitute the limits and integrate with respect to x: ∫ (-3x^3/2 + 6x^2 + 2x) dx, with the limits of integration 0 and 3.
9. Evaluate the integral: (-3(3)^3/2 + 6(3)^2 + 2(3)) - (-3(0)^3/2 + 6(0)^2 + 2(0)) = 19.5.
10. The final answer is 19.5 units of upward flux.