# Using triple integrals to find volume

Find the volume of the solid defined by the inequalities

$0 \le z \le y \le x \le 1$.

I know I have to use triple integrals to solve this problem, but I am pretty confused as to how I should approach it. I am having trouble figuring out what the limits of integration are for $x$,$y$, and $z$ given the above inequality.

• Isn't this a cube with sides just less than1? – JTP - Apologise to Monica Jul 21 '15 at 1:29
• Hint: the region in the $xy$ plane bounded by $y=0, y=x, x=1$ is the "floor" of the solid, and is also the region of projection. The "roof" is the plane $z=y$. – Ned Jul 21 '15 at 1:36

## 1 Answer

If you were to do this with integrals, where $S = \{ (x,y,z) : 0 \le z \le y \le x \le 1 \}$ your integral limits would be $$\text{Volume} = \iiint_S dV = \int_0 ^1 \int_0 ^x \int_0 ^y dz dy dx = \frac{1}{6}$$