Integration over the cube

Question

I have the following exercise:

Integrate the $g=x \cdot y \cdot z$ over the cube that is on the first octant and that is bounded from the levels $x=1, y=1, z=1$.

Could you give me some hint what I am supposed to do?

I got stuck right now...

I have the following formula for the surface integral in my notes: $$z=f(x,y)$$ $$\iint_R g(x,y,z) d \sigma=\iint_S g(x,y,z) \sqrt{f_x^2+f_y^2+1}dxdy$$ How can I apply this formula in this case?

Answer 1

Your formula is to integrate a function over a surface. It does not apply in this case since you're integrating over a volume instead of over a surface.

You are supposed to multiply your function with an infinitesimal volume element and integrate over the octant. That is: $$\iiint_{Octant} g(x,y,z) dV = \iiint_{Octant} xyz\ dxdydz$$

Answer 2

$$ \iiint \limits_V xyzdV=\int_0^1\int_0^1\int_0^1xyzdxdydz. $$