Evaluate $$\iiint_E xyz\, dV$$ where $E$ is the solid: $0\leq z\leq 9,\,0\leq y\leq z,\, 0\leq x \leq y.$

I am having a hard time drawing a picture of this solid $E$ to find out what the limits of integration can be. I see when $x=0$ the solid in the $yz$-plane is a triangle and then as we move down the positive $x$-axis for $0\leq x \leq 9$, the triangle gets smaller and converges to a point when $x=9$. That is about all I have. I don't know how to visualize this perfectly enough in my mind and hold it steady to see which direction I should integrate in first and then find out what are the equations of the planes that I need to set up the integral. Any help is greatly appreciated.

  • $\begingroup$ Integrate for $x$ from $0$ to $y$, then integrate for $y$ from $0$ to $z$ and finally integrate for $z$. The body is a triangular pyramid. $\endgroup$
    – Džuris
    Commented Apr 18, 2014 at 19:25
  • 1
    $\begingroup$ This was part of my now deleted answer that got a downvote\begin{eqnarray*} I &=&\iiint_{E}xyz\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}xyz\,dx\,dy \,dz \\ &=&\int_{0}^{9}\,dz\int_{0}^{z}dy\int_{0}^{y}xyz\,dx \\ &=&\int_{0}^{9}\left( \int_{0}^{z}\left( \int_{0}^{y}xyz\,dx\right) \,dy\right) \,dz \\ &=&\dots\\ &=&\left. \frac{1}{48}z^{6}\right\vert _{0}^{9} \\ &=&\frac{531\,441}{48}=\frac{177\,147}{16}. \end{eqnarray*} $\endgroup$ Commented Apr 18, 2014 at 19:28
  • $\begingroup$ @AméricoTavares I did not downvote but it seemed wrong. If we'd take shape where all coordinates are between 0 and 9 we can't get bigger than a cube with side length 9 which has a volume of 729. $\endgroup$
    – Džuris
    Commented Apr 18, 2014 at 19:29
  • $\begingroup$ @Juris Thanks for your comment but note that we are evaluating $$\iiint_{E}xyz\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}xyz\,dx\,dy \,dz$$ and not $$\iiint_{E}\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}\,dx\,dy \,dz$$ $\endgroup$ Commented Apr 18, 2014 at 19:33
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    $\begingroup$ My bad, I agree with your result then :) $\endgroup$
    – Džuris
    Commented Apr 18, 2014 at 19:35

2 Answers 2


You can find the limits of integration by solving the inequalities. But in this case, the inequalities are already in solved form, if you integrate in the order $x,y,z$ (listed from innermost to outermost), so there's nothing to do.

If you wanted to integrate in the order $z,y,x$, we could solve in this other direction. In this case, it's pretty easy: we can combine all of the inequalities into one

$$ 0 \leq x \leq y \leq z \leq 9 $$

and then just break it apart in the other direction:

$$ 0 \leq x \leq 9 \qquad x \leq y \leq 9 \qquad y \leq z \leq 9 $$

Or if we wanted $x,z,y$, we could use

$$ 0 \leq y \leq 9 \qquad 0 \leq x \leq y \qquad y \leq z \leq 9 $$

This last version is neat, because the two inner integrals over $x$ and $z$ are actually independent of each other, and you can save yourself a bit of trouble by factoring the double integral into a product of two single integrals.

It's actually okay to get things too wide as you're working, as long as you don't throw away information in your inequalities: you'll correct yourself later on.

If, in the last example, I determined that $y$ was restricted to $0 \leq y \leq 10$, then later on when I moved to solve for $z$, I realize I need to split $y \leq z \leq 9$ into the $y \leq 9$ and $y \geq 9$ cases in order to properly write down bounds on my integrals and will quickly realize that there is nothing to integrate in the $9 \leq y \leq 10$ region. (this was a little obvious in this case, but the actual extents of your variables more complicated system of inequalities can take more effort to pin down exactly)


enter image description here

You might think about a diagram like this: region between the white planes for x, region between the pink planes for y, region between the cyan planes for z.

The solid is a bit easier to see now, demarcated with blue points


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