Figuring the order of integration

Question

I have this function $f(x,y,z)=1/3(3x+2y+z)I(0<x<1)I(0<y<1)I(0<z<1)$ and want to integrate over this region s.t. $x>1/2$, $y<1/2$, and $z < 1/2$. So this will be a triple integral. How do I know what ORDER to integrate?

i.e, $dzdydx$? $dxdydz$ ?

Answer 1

Any order you like. All of them will give the same result. Because this definite integral actually represents the volume enclosed by a closed surface, nothing else.

As for example: $f(x,y,z)=\frac{1}{3}(3x+2y+z)$ $$\int f dV = \int \int \int \frac{1}{3}(3x+2y+z) dx dy dz$$ Here x,y,z are not linked by any function i.e. the restriction on x,y,z are independent, not via any function. So we put the limits directly. $$\int f dV = \int_{x=\frac{1}{2}}^{x=1} \int_{y=0}^{y=\frac{1}{2}} \int_{z=0}^{z=\frac{1}{2}} \frac{1}{3}(3x+2y+z) dx dy dz$$