Question

高等数学，第二题，怎么做。要详细步骤

Answer 1

I = ∫∫ <D>x√(1-y^2)dxdy = ∫<下0, 上1>xdx ∫<0, x> √(1-y^2)dy
= ∫<0,1>xdx [(1/2)y√(1-y^2)+(1/2)arcsiny]<0, x>
= (1/2) ∫<0,1> [x^2√(1-x^2)+xarcsinx]dx
令 x=sint， 得
I = (1/2) ∫<0,π/2> [(sint)^2cost+tsint]costdt
= (1/8) ∫<0,π/2> (sin2t)^2dt + (1/4) ∫<0,π/2> tsin2t dt
= (1/16) ∫<0,π/2>(1- cos4t)dt - (1/8) ∫<0,π/2> t dcos2t
= (1/16) [t-(1/4)sin4t]<0,π/2> - (1/8) [tcos2t-(1/2)sin2t<0,π/2>
= π/32 + π/16 = 3π/32