Question

Let F(x, y, z) = 3xi+ 2yj and let σ be the cube with opposite corners at (0, 0, 0) and (5, 5, 5), oriented outwards. Find the flux of the flow field F across σ.

Answer 1

To find the flux of the flow field F across the cube σ, we need to calculate the surface integral of the dot product of F and the outward unit normal vector of σ. Divide the cube into six individual faces and calculate the flux through each face separately. The net flux can be found by summing up the flux through each face.

Step-by-step explanation:

To find the flux of the flow field F across σ, we need to calculate the surface integral of the dot product of F and the outward unit normal vector of σ. Since σ is a cube with opposite corners at (0, 0, 0) and (5, 5, 5), we can divide it into six individual faces and calculate the flux through each face separately.

Let's denote the faces of the cube as A, B, C, D, E, and F. The net flux can be found by summing up the flux through each face.

Flux through face A: Φ_A = ∫∫_A (F · dS) = ∫∫_A (3x · dS) = 3 ∫∫_A x dS

Similarly, we can calculate the flux through faces B, C, D, E, and F using the same process. After finding the flux through each face, the net flux can be calculated by adding them together: Φ_net = Φ_A + Φ_B + Φ_C + Φ_D + Φ_E + Φ_F

Answer 2

Use the divergence theorem,

`\displaystyle\iint_(\partial\sigma)\mathbf F(x,y,z)\cdot\mathrm d\mathbf S=\iiint_\sigma\mathrm{div}\mathbf F(x,y,z)\,\mathrm dV`

We have

`\mathrm{div}\mathbf F(x,y,z)=(\partial(3x))/(\partial x)+(\partial(2y))/(\partial y)+(\partial0)/(\partial z)=5`

so that the flux across
`\sigma` is equal to 5 times the volume of the cube. The cube itself has edge length 5, so its volume is
`5^3=125`, making the flux
`5^4=\boxed{625}`.