Question

求曲线y=√(3-t^2)在-√3到x的定积分的全长

Answer 1

y=∫(-√3,x)√3-t^2dt

ds=√1＋y'平方 dx

=√1+3-x平方dx

=√4-x平方dx

s=∫(-√3,√3)√4-x平方dx

令x=2sint

原式=2∫(0,√3)√4-x平方dx

=4∫(0,π/3)2cos平方tdt

=4∫(0,π/3)（1+cos2t）dt

=4(t+1/2 sin2t)|(0,π/3)

=4(π/3+√3/4-0)

=4π/3+√3