Evaluate the triple integral of xyz over the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Step-by-step solution

1. 1

   Check that the integral separates. All six limits are constants and the integrand is a pure product $x \cdot y \cdot z$, with each factor depending on only one variable. That is exactly the condition under which a multiple integral factors into a product of single integrals — the shortcut used at the end as a check.

2. 2

   Integrate in z, treating x and y as constants.

   $\int_0^1 xyz\,dz = xy\left[\frac{z^2}{2}\right]_0^1 = \frac{xy}{2}$

   Only $z$ is being integrated, so $xy$ rides along untouched.

3. 3

   Integrate in y.

   $\int_0^1 \frac{xy}{2}\,dy = \frac{x}{2}\left[\frac{y^2}{2}\right]_0^1 = \frac{x}{4}$

4. 4

   Integrate in x.

   $\int_0^1 \frac{x}{4}\,dx = \frac{1}{4}\left[\frac{x^2}{2}\right]_0^1 = \frac{1}{8}$

5. 5

   Confirm with the separated form. Because the integral factors,

   $\int_0^1 \!\! \int_0^1 \!\! \int_0^1 xyz\,dV = \left(\int_0^1 x\,dx\right)\left(\int_0^1 y\,dy\right)\left(\int_0^1 z\,dz\right) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$

   This is also the mean of $xyz$ over the cube (the cube has volume $1$), which matches the fact that $x$, $y$, $z$ chosen independently and uniformly have expected product $\tfrac12 \cdot \tfrac12 \cdot \tfrac12$.