Question

求二重积分∫∫（y√1+x^2-y^2）dt,其中D是由直线y=x,x=-1和y=1所为成的闭区域

Answer 1

简单计算一下即可，答案如图所示

Answer 2

本题需要先积y,若先积x计算量会很大.
∫∫（y√1+x²-y²）dxdy
=∫[-1--->1] dx ∫[x--->1]（y√1+x²-y²）dy
=(1/2)∫[-1--->1] dx ∫[x--->1]（√1+x²-y²）d(y²)
=(-1/2)∫[-1--->1] (2/3)(1+x²-y²)^(3/2) |[x--->1] dx
=(-1/3)∫[-1--->1] [|x|³-1] dx 注意这里不能写x³,因为x有负值
被积函数是偶函数,由奇偶对称性
=(-2/3)∫[0--->1] [|x|³-1] dx
=(2/3)∫[0--->1] [1-x³] dx
=(2/3)(x-x⁴/4) |[0--->1]
=(2/3)(1-1/4)
=1/2