A1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. its resistivity depends on the distance x from the left end and obeys the formula rho( x ) = a + bx^2, where a and b are constants. at the left end, the resistivity is 2.25 10^-8 omega*m, while at the right end it is 8.50 x 10^-8 omega*m.

what is the resistance of this rod?

what is the electric field at its midpoint if it carries a 1.75 a current?

if we cut the rod into two 75.0 cm halves, what is the resistance of each half?

I have solved parts A and B by doing the following:
A) R = ρL/A First I found the constants a and b by solving for them, which I found to be: a = 2.25x10^-8 Ωm b = 2.78x10^-8 Ωm A = πr^2 A = 3.8013x10-4 m2 Taking the integral of the Resistance equation gave me: 1/A*(ax + bx3/3) I then took the integral from 0 to 1.50m, thereby giving me: (1/3.801299x10-4)*3.38x10-8*1.50 + 2.78x10-8*1.53/3 = 1.71x10-4 This I know to be correct.
B) ∫E.dl = V dV/dx = d/dx(IR) 1.75/A*(a + bx2) Plugging in 0.75 for x gave me E = 1.76x10^-4 V/m This is also correct
C) This is where I'm stuck. I had thought that if I used the integral in part A to find the integral from 0 to 0.75 then that would be my resistance in the left hand side. Then I thought that I could subtract the left hand value from the value I found in part A to give me the right hand value, but this hasn't worked. My workings are as follow: R = ρL/A A = 3.8x10^-4 m2 Rleft hand1/A*(2.25x10^-8)(0.75) + (2.78x10^-8)(0.75^3/3) = 4.43966x10^-5 Then Rright hand = 1.71x10^-4 - 4.43966x10^-5 = 1.266x10^-4 

Full change in the electric plus system