将∫(0,1)dx∫(0,1-x)dy∫(0,x+y)f(x,y,z)dz按y,z,x的次序积分为?
将∫(0,1)dx∫(0,1-x)dy∫(0,x+y)f(x,y,z)dz按y,z,x的次序积分为?
变换积分次序∫(0,1)dy∫(-y,1+y^2)f(x,y)dx
高数重积分的问题∫(0→1)dx∫(0→x)dy∫(0→y)f(z)dz=1/2∫(0→1)(1-z)^2f(z)dz
f(x,y,z)=0,z=g(x,y),求dy/dx,dz/dx
交换积分次序∫(1,0)dx∫(x,0)f(x,y)dy
∫[0,1] dx∫[-x^2,1] f(x,y)dy交换积分次序
∫(L的换积分)(y-z)dx+(z-x)dy+(x-y)dz,L为x^2+y^2+z^2=1与(x-1)^2+(y-1
交换累次积分的次序∫(0>1) dy∫(0>2y) f(x,y)dx +∫(1>3) dy∫(0>3-y) f(x,y)
交换积分次序∫(1,0)dx∫(x,0)f(x,y)dy+∫(2,1)dx∫(2-x,0)f(x,y)dy
交换积分次序∫(0,1)dy∫(0,y)f(x,y)dx+∫(1,2)dy∫(0,2-y)dxf(x,y)dx
交换积分次序 ∫(4,0)dx∫(x,2x^0.5)f(x,y)dy
更换积分次序∫(0,2)dx∫(x,3x)f(x,y)dy