高等数学曲线积分题目求解

Question

如题所述

Answer 1

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Answer 2

7 M = ∫<L>√(2y/a)ds = a√ ∫<0,1> t√(1+t^2+t^4)dt (u=t^2)
= (a/2)√ ∫<0,1> √(1+u+u^2)du
= (a/2)√ ∫<0,1> √[3/4+(u+1/2)^2]d(u+1/2)
= (a/4)【(u+1/2)√[3/4+(u+1/2)^2]+(3/4)ln{u+1/2+√[3/4+(u+1/2)^2]}】<0,1>
= (a/4)[(u+1/2)√(1+u+u^2)+(3/4)ln{u+1/2+√(1+u+u^2)}]<0,1>
= (a/4)[(3/2)√3-1/2+(3/4)ln(3/2+√3)-(3/4)ln(1/2+1)]
= (a/4)[(3/2)√3-1/2+(3/4)ln(1+2√3/3)]

Answer 3

曲线的线密度 u=t
质量
m =∫{0,1}tds = ∫{0,1}t(a^2+(at)^2+(at^2)^2)^(1/2)dt
=（a/2）∫{0,1}(1+u+u^2)^(1/2)du {这里u=t^2}
=(a/4)*[(u+1/2)(u^2+u+1)^(1/2)+(3/4)*ln(u+1/2+(u^2+u+1)^1/2)]_{0,1}
=(a/4)*[(3/2)*√3 +(3/4)*ln((3/2)+√3) -(1/2)-(3/4)*ln(3/2) ]