# Evaluate $\iiint xyz$

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.

• 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. Commented Apr 18, 2014 at 19:25
• 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*} Commented Apr 18, 2014 at 19:28
• @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. Commented Apr 18, 2014 at 19:29
• @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 $$\iiint_{E}\,dV=\int_{z=0}^{9}\int_{y=0}^{z}\int_{x=0}^{y}\,dx\,dy \,dz$$ Commented Apr 18, 2014 at 19:33
• My bad, I agree with your result then :) Commented Apr 18, 2014 at 19:35

## 2 Answers

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)

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