Considering the side view of the given solid showing trapezium section.The mid point of rectangular surface #A# is considered as origin O and #OX# ,the X-axis is perpendicular to surface #A# at O.
We also consider a rectangular slice of length #y# and width #w# parallel to the surface #A# at a distance #x# from origin #O#. The slice has infinitesimally small thickness #dx#.
Now length of the rectangular cross section increases from A to B at a definite rate along X-axis and the rate is#=(y_2-y_1)/L#
So #y=y_1+x((y_2-y_1)/L)#
Hence area of the selected slice
#A=wxxy#
If #dR# be the resistance of the slice then we can write
#dR=rho(dx)/A=rho(dx)/(wy)=rho/w((dx)/(y_1+x((y_2-y_1)/L)))#
Taking integration between #(0,L)# we get the resistance of the solid across AB
#R=int_0^Lrho/w((dx)/(y_1+x((y_2-y_1)/L)))#
#=>R=rho/wint_0^L((dx)/(y_1+x((y_2-y_1)/L)))#
#=>R=rho/wxxL/(y_2-y_1)[ln(y_1+x((y_2-y_1)/L))]_0^L#
#=>R=rho/wxxL/(y_2-y_1)[ln(y_1+L((y_2-y_1)/L))-ln(y_1+0xx((y_2-y_1)/L))]#
#=>R=rhoL/(w(y_2-y_1))ln(y_2/y_1)#
