Net Volume/Mass of Fluid Through Surface

[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.

Homework Statement

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.

Homework Equations

[tex]\Phi = \int\int F \cdot (r_u \times r_v) dA [/tex]

The Attempt at a Solution

First to parametrize the surface:
x = u; y = v; z = 1 - u - v
[tex]r_u = <1, 0, -1> [/tex]
[tex]r_v = <0, 1, -1> [/tex]
[tex]r_u \times r_v = <1,1,1>[/tex]

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

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?

 

[mighty2000]

Hello,

Thank you for sharing your solution. It looks like you have correctly parametrized the surface and set up the integral. I followed your steps and got the same result of 0 m^3. However, I believe the mistake may be in the given answer in the back of the textbook. Based on the given velocity vector, it seems like the fluid particle is only moving in the x and y direction, not the z direction. Therefore, the z component of the velocity vector should be 0, not z. This would change the integral to:

\Phi = \int\int <2u,-3v,0> \cdot <1,1,1> dA
=\int\int (2u - 3v) dA
=\int_0^1 \int_0^{1-u} (2 + 2u - 3v) dv du
=\int_0^1 {(2v + v^2 + 2uv)}_0^{1-u} du
=\int_0^1 ((2 + u + 2u - 3u^2)) du
=\int_0^1 (2 + 3u - 3u^2) du

which gives a net volume of 32/3 m^3 and a net mass of 32/3*806 = 25792/3 kg. I hope this helps and please let me know if you have any other questions.