# Flux Integral

1. Sep 29, 2008

### Knissp

EDIT: NEVER MIND, I got 32/3 for the very next question, so I'm pretty sure the answer that I was comparing this to from the back of my textbook has a typo for the question number. If anyone still wants to check this, that would be OK but I'm pretty sure it's correct.

1. The problem statement, all variables and given/known data
Let F(x,y,z) = <2x,-3y,z> be the velocity vector (in m/s) of a fluid particle at the point (x,y,z) in steady state fluid flow.
a) Find the net volume of fluid that passes in the upward firection through the portion of the plane x+y+z=1 in the first octant in 1 s.
b) Assuming that the fluid has a mass density of 806 kg/m^3, find the net mass of fluid that passes in the upward direction through the surface in part (a) in 1 s.

2. Relevant equations
$$\Phi = \int\int F \cdot (r_u \times r_v) dA$$

3. The attempt at a solution
First to parametrize the surface:
x = u; y = v; z = 1 - u - v
$$r_u = <1, 0, -1>$$
$$r_v = <0, 1, -1>$$
$$r_u \times r_v = <1,1,1>$$

$$\Phi = \int\int F \cdot (r_u \times r_v) dA$$
$$=\int\int <2u,-3v,(1-u-v)> \cdot <1,1,1> dA$$
$$=\int\int (2u - 3v + 1 - u - v) dA$$
$$=\int_0^1 \int_0^{1-u} (1 + u - 4v) dv du$$
$$=\int_0^1 {(v + uv - 2v^2)}_0^{1-u} du$$
$$=\int_0^1 ((1-u) + u(1-u) - 2(1-u)^2) du$$
$$=\int_0^1 (-1-3u^2+4u) du$$

which equals 0 m^3. This means that the net mass of the fluid that passes through the surface is also 0 kg. BUT the back of the textbook says the answer to part (b) is 32/3. Did I mess up somewhere?