Introduction: In this problem, we are given a cylindrical surface sheet with radius r=20m, ø=55, and z=5m. We are required to determine the flux crossing 1mm by 1mm area on this surface sheet if D =4Xāx 2(1-y)āy 4zāz C/m2.

Explanation: To determine the flux crossing the given surface sheet, we first need to find the electric field intensity E. We can use the relation E = D/ε, where ε is the permittivity of free space. The value of ε is 8.854 X 10^-12 F/m.

Calculation: Substituting the given values of D and ε, we get E = (4Xāx 2(1-y)āy 4zāz C/m2)/(8.854 X 10^-12 F/m) = 4.51 X 10^11 (āx 2(1-y)āy 4zāz) V/m.

Now, we can use the formula for flux, Φ = ∫∫ E.dS, where dS is the area element. Since the surface sheet is cylindrical, we can use cylindrical coordinates (r, ø, z) to express the area element as dS = r.dr.dø.

We are given that the area over which we need to find the flux is 1mm by 1mm. Converting this to meters, we get 0.001m by 0.001m. We can use this area to calculate the limits of integration for r and ø.

The limits of integration for r are r = 20m (the radius of the cylindrical surface sheet) and r = 20.001m (adding 1mm to the radius). The limits of integration for ø are ø = 55° and ø = 56° (adding 1mm to the angle). The limits of integration for z are z = 5m and z = 5.001m (adding 1mm to the height).

Final Answer: Substituting the limits of integration and the expression for E, we get Φ = ∫∫∫ E.dS = ∫20m^20.001m∫55°56°∫5m^5.001m 4.51 X 10^11 (āx 2(1-y)āy 4zāz) r.dr.dø.dz = 2.26 X 10^-3 C.

Therefore, the flux crossing the 1mm by 1mm area on the cylindrical surface sheet is 2.26 X 10^-3 C.