If D = 2xy i + 3yz j + 4xz k, how much flux passes through x = 3 plane for which -1<y<2 and 0<z<4?

To find the flux ( Phi ) of the vector field ( D = 2xy hat{i} + 3yz hat{j} + 4xz hat{k} ) through the plane ( x = 3 ), we can use the formula for flux:

[

Phi = iint_S mathbf{D} cdot mathbf{n} , dS

]

where ( mathbf{n} ) is the outward normal vector to the surface ( S ), and ( dS ) is the differential area element.

1. Identify the normal vector: For the plane ( x = 3 ), the outward normal vector is ( mathbf{n} = hat{i} ).

2. Evaluate the vector field on the plane: Since we are considering the plane where ( x = 3 ), we substitute ( x = 3 ) into the vector field ( D ):

[

D = 2(3)y hat{i} + 3yz hat{j} + 4(3)z hat{k} = 6y hat{i} + 3yz hat{j} + 12z hat{k}

]

3. Take the dot product with the normal vector:

[

mathbf{D} cdot mathbf{n} = (6y hat