y=5±√(16-x^2)
V=π∫(-1,1)((5+√(16-x^2))-(5-√(16-x^2)))dx
V=2π∫(-1,1)√(16-x^2)dx
V=4π∫(0,1)√(16-x^2)dx
先抛开积分x的上下限
设x=4sint,t属于(-π/2,π/2)
sint=x/4
cost=√(16-x^2)/4
V=4π∫4costd(4sint)
V=64π∫(cost)^2dt
V=32π∫[cos(2t)+1]dt
V=16π∫cos(2t)d(2t)+32π∫1dt
V=16πsin(2t)+32πt
V=32π(sintcost+t)
V=32π[x√(16-x^2)/16+arcsin(x/4)]
V=2πx√(16-x^2)+32πarcsin(x/4)
代入积分上下限(0,1)
得V=2π√15+32πarcsin(1/4)