Question

find the volume solid generated by revolution of candiod curve

Answer 1

The parametric equations of a cardioid are x=cosθ(1−cosθ)x=cos⁡θ(1−cos⁡θ) and y=sinθ(1−cosθ)y=sin⁡θ(1−cos⁡θ), 0≤θ≤2π0≤θ≤2π. Diagram here. The region enclosed by the cardioid is rotated about the x-axis, find the volume of the solid. I am not allowed to use polar form, or double integrals, due to the limitations of the NSW mathematics syllabus. Basically I'm stuck with using a disk approximation. What I've got so far using the disk approximation is:

limδx→0∑0x=−2πy2δx=π∫0−2y2dx=π∫π2πsin2θ(1−cosθ)2(2cosθsinθ−sinθ)dθlimδx→0∑x=−20πy2δx=π∫−20y2dx=π∫ππ2sin2⁡θ(1−cos⁡θ)2(2cos⁡θsin⁡θ−sin⁡θ)dθwhich gives the volume of the portion from x= -2 to 0, but I have no idea how to calculate the volume from x = 0 to 1/4. My gut feeling says to just extend the bounds of the previous integral to ππ and 0, but because there exists two y-values for each x value in that domain, shouldn't I have to subtract the volume of the larger disk from the smaller disk i.e. form an annulus?