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 solve this, we’ll use the concept of flux through a surface. The flux (Phi) of a vector field (textbf{D} = Ptextbf{i} + Qtextbf{j} + Rtextbf{k}) through a surface (S) is given by the surface integral of (textbf{D} cdot textbf{n} dS), where (textbf{n}) is the unit normal to the surface and (dS) is a differential element of the surface area.

Given (textbf{D} = 2xy textbf{i} + 3yz textbf{j} + 4xz textbf{k}) and considering the plane (x = 3) with (-1 < y < 2) and (0 < z < 4), we'll calculate the flux through this plane area.

For the plane (x = 3), the normal vector is parallel to the (textbf{i}) direction since the plane is perpendicular to the x-axis. Therefore, only the component of (textbf{D}) in the direction of (textbf{i}) contributes to the flux through this plane.

The relevant component of (textbf{D}) here is (P = 2xy), and since (x = 3), we have (P = 6y). The