Question

求二重积分：∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)dy 疑问： 解法一：∫[-1,1]dx∫[0,...

求二重积分：∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)dy
疑问：
解法一：∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)dy=2∫[0,1]dx∫[0,x^2](x^2-y)^(1/2)dy=4/3∫[0,1]x^3dx=1/3
解法二：∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)dy=2/3∫[-1,1]x^3dx=0
请解释谁对谁错。
注：[*，*]为[积分下限，积分上限]。
一楼说的是第一种解法，能否说说第二种错在哪里

Answer 1

两种解法都是错误的
∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)dy
＝-∫[-1,1]dx∫[0,x^2](x^2-y)^(1/2)d(x^2-y)
=-2/3∫[-1,1]dx(x^2-y)^(3/2)[0,x^2]
=2/3∫[-1,1]|x^3|dx
=4/3∫[0,1]x^3dx
=1/3x^4[0,1]
=1/3