Question

求高数大神解答，谢谢😃

Answer 1

x^2+y^2+z^2=2∫(x->y) f(x+y-t) dt

let
u= x+y-t
du = -dt
t=x, u = y
t=y, u=x

∫(x->y) f(x+y-t) dt
=∫(y->x) f(u) (-du)
=∫(x->y) f(t) dt

x^2+y^2+z^2=2∫(x->y) f(x+y-t) dt
x^2+y^2+z^2=2∫(x->y) f(t) dt
2x + 2z. ∂z/∂x = -2f(x)
z. ∂z/∂x = -x-f(x)

Similarly

x^2+y^2+z^2=2∫(x->y) f(x+y-t) dt
x^2+y^2+z^2=2∫(x->y) f(t) dt
2y + 2z. ∂z/∂y = 2f(y)
z. ∂z/∂y = -y+f(y)

z. (∂z/∂x +∂z/∂y)
=-x-f(x) -y+f(y)