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Evaluate the triple integral:

$$\iiint_E xyz \; \mathrm dV$$

where $E$ is the solid: $0\le z\le 2, 0\le y\le z, 0\le x\le y$.

Please explain this, I'm so confused.

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  • $\begingroup$ $$\int_0^2\int_0^z\int_0^y (xyz)dxdydz$$ $\endgroup$ – anon Mar 20 '13 at 21:42
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If you think of $dx$ as a change in distance, then $dydx$ would be (a change in distance) * (a change in distance) = (a little change in distance)^2 = (a little change in area). One more and it's a change in volume.

$dV$ = change in volume = $dxdydz$

$\int\int\int_{E}(xyz)dxdydz$

E = the volume in question, but you already have the bounds. Note that for this integral you need to integrate in a specific order. $x$ is evaluated from $0$ to $y$, so it should be evaluated before $y$ is. Otherwise you'll be left with a $y$ in your final answer. If we follow this pattern, we can see that $x$ must be integrated before $z$. That leaves $z$ to be integrated between two numbers, ultimately leaving you with the answer you want (ie a number).

$\int_{0}^{2}\int_{0}^{z}\int_{0}^{y}(xyz)dxdydz$

Hope that helps.

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In general, one can compute integrals of functions over regions to get numbers. For instance, one can integrate a function which is a function of three variables over a three-dimensional region, or a volume. This is what you're doing. You're integrating the function $f(x,y,z) = xyz$ over the region $E$.

What might be giving you a hangup is that the bounds of integration involve more variables. The way to approach that is to put in the bounds of integration in, making sure that you put the ones involving variables on the inside. Otherwise you wouldn't get a number at the end. The end result is just as anon wrote it in the comments.

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