根号下(x^2-y^2)在(0,x)对y求积分怎么算？

Question

如题所述

Answer 1

对y求不定积分,即当x为常数:
∫[0,x]√(x²-y²)dy
令y=xsinp,dy=xcospdp
当y=0,p=0,当y=x,p=π/2
=∫[0,π/2]xcosp·√(x²-x²sin²p) dp
=x∫[0,π/2]cosp·xcosp dp
=x²∫[0,π/2]cos²p dp
=x²∫[0,π/2](1/2)(1+cos2p) dp
=(x²/2)∫[0,π/2]dp+(x²/2)∫[0,π/2]cos2pdp
=(x²/2)p[0,π/2]+(x²/2)(1/2)sin2p[0,π/2]
=(x²/2)(π/2)+(x²/4)(0)
=x²π/4