
Bradford T. answered 05/15/21
MS in Electrical Engineering with 40+ years as an Engineer
The initial line is the x-axis
S = 2π∫0π y ds
ds = √(r2+(r')2 dθ = √(a2(1+cosθ)2 + (-asinθ)2)dθ = a√(2(1+cosθ))dθ
S = 2πa2∫0π (1+cosθ)sinθ√2 √(1+cosθ) dθ
let u = 1+cosθ, du = -sinθ
when θ= 0, u = 2
when θ= π, u=0
S = -2√2 πa2∫20u3/2du = 2√2 πa2∫02u3/2du = 2√2(2/5)a2πu5/2|02
S = 4√2πa225/2/5 = 32a2/5
Emmma W.
Can u give it in hand written?05/15/21