Question

高等数学计算问题，画红线的地方怎么算出的，求详细过程

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Answer 1

左处：... = (1/2) ∫ <0, 1> dx [(2/3)(1-x^2+y^2)^(3/2)]<0, x>
= (1/3) ∫ <0, 1> [1-(1-x^2)^(3/2)]dx = ...
右处：... = ∫<1, 2>dy∫<y, y^2> (2y/π)sin[πx/(2y)]d[πx/(2y)]
= (2/π)∫<1, 2>ydy[-cos(πx/2y)]<y, y^2>
= (2/π)∫<1, 2>y[cos(π/2)-cos(πy/2)]dy
= (-2/π)∫<1, 2>ycos(πy/2)dy = ...