此题用直角坐标求解简单些!
原式=∫(0,1)dx∫((1-x),√(1-x²))ydy (符号“∫(a,b)”表示从a到b积分)
=1/2∫(0,1)dx∫((1-x),√(1-x²))d(y²)
=1/2∫(0,1)[(1-x²)-(1-x)²]dx
=1/2∫(0,1)(2x-2x²)dx
=∫(0,1)(x-x²)dx
=(x²/2-x³/3)|(0,1)
=1/2-1/3
=1/6
此题用直角坐标求解简单些!
原式=∫(0,1)dx∫((1-x),√(1-x²))ydy (符号“∫(a,b)”表示从a到b积分)
=1/2∫(0,1)dx∫((1-x),√(1-x²))d(y²)
=1/2∫(0,1)[(1-x²)-(1-x)²]dx
=1/2∫(0,1)(2x-2x²)dx
=∫(0,1)(x-x²)dx
=(x²/2-x³/3)|(0,1)
=1/2-1/3
=1/6