Question

√x+√y=1的曲线的弧长，运用定积分方法求

Answer 1

由 √x+√y=1, 得 √y=1-√x, y'/(2√y)=-1/(2√x), 则 y'=-√y/√x=-(1-√x)/√x=1-1/√x.
L =∫<0,1>√(1+y'^2)dx =∫<0,1>√[1+(1-1/√x)^2]dx
=∫<0,1>√(2-2/√x+1/x)dx, 令 √x=t， 则 x=t^2, dx=2tdt, 得
L=∫<0,1>√(2-2/t+1/t^2)2tdt = 2∫<0,1>√(2t^2-2t+1)dt
= 2√2∫<0,1>√[(t-1/2)^2+1/4]d(t-1/2) 令u=t-1/2, 得
L = 2√2∫<-1/2,1/2>√(u^2+1/4)du = 4√2∫<0,1/2>√(u^2+1/4)du
= 2√2[u√(u^2+1/4)+(1/4)ln{u+√(u^2+1/4)}]<0,1/2>
= 1+(1/√2)ln(1+√2).

Answer 2

解:
0<=x<=1
0<=y<=1
∫[0,1]∫[0,1](根号x +根号y)dxdy
=∫[0,1]dx∫[0,1](根号x +根号y)dy
=∫[0,1]dx(2/3y^3/2)|[0,1]
=∫[0,1]2/3dx
=2/3