compute the surface area of the surface generated by revolving one arch of the cycloid c(t) = (8t-8sint,8cost) about the x-axis

Answer 1

dSA=2π*y(t)*ds

y(t)=8(1-cos(t)

dx/dt=8(1-cos(t))

dy/dt=8sin(t)

(ds/dt)²=(dx/dt)²+(dy/dt)²=64(2-2cos(t)),,,,,note: 2cos(t)=2(1-2sin²(t/2))

ds=16sin(t/2)dt

SA=32π*∫(8(1-cos(t))sin(t/2)dt,,,,,t=0 to 2π

1-cos(t)=2-2cos²(t/2)

SA=256π*∫[2sin(t/2)-2cos²(t/2)sin(t/2)]dt,,,t=0 to 2π

careful that for cos²(t/2), that the derivative of cos(t/2) is -(1/2)sin(t/2)

SA=512π(-2cos(t/2)+(2/3)cos³(t/2),,,,t=0 to 2π

SA=512π*(8/3)