Have a problems with the following question:

Consider the solid of revolution obtained when the region surrounded by x=3 ,x=6 ,y=0 and y= x+5 is rotated about the line y=2

Enter an expression for the cross-sectional area of this solid at a point

I can find the value..but how to get the expression?

I reckon the ans is suppose to be

((x+3)^2)*pi

but its wrong