计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__________.

(int )_(0)^1dx(int )_(x)^1(e)^-(y^2)dy= () .-|||-

把下列积分化为极坐标形式, 并计算积分值: (3)(int )_(0)^1dx(int )_({x)^2}(({x)^2+(y)^2)}^-df

24.证明: (int )_(0)^1dx(int )_(0)^xf(x-y)dy=(int )_(0)^1f(x)(1-x)dx-|||-"f(x-y)dy=

在下列积分中改变累次积分的顺序:-|||-(1) (int )_(0)^2dx(int )_({x)^2x}f(x,y)dy;-|||-(2) (int )_(

int dfrac (1)({x)^2}sqrt (dfrac {1-x)(1+x)}dx

[例5] 积分 (int )_(0)^2dx(int )_(0)^sqrt (2x-{x^2)}sqrt ({x)^2+(y)^2}dy= __

计算 =(int )_(L)dfrac ((x-y)dx+(x+y)dy)({x)^2+(y)^2}-|||-f[(x-y)(x+(x+y)dy)/(x^2+x

12.化下列二次积分为极坐标形式的二次积分:-|||-(3) (int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}f(x,y)dy

(6) () =dfrac (1)(sqrt {1-{x)^2}} int dfrac (1)(sqrt {1-{x)^2}}dx=() .

(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1