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I need to change order of integration for the following triple integral: $$ \int\limits_{0}^{2} \int\limits_{0}^{2z} \int\limits_{y}^{2y} f_{(x,y,z)}\; dx\, dy\, dz $$

The domain of integration is graphically described in the following images:

Y-Z plane

enter image description here

X-Y plane

enter image description here

By setting $f_{(x,y,z)} = 1$, I verified my results, comparing them to the original integral using WA, which yields $\frac{16}{3} \cong 5.33$.

The requirement

What I'm trying to do is to change the order of integration, so that:

  1. The volume will be projected onto the X-Z plane
  2. Having x to be a function of z (or vice versa - z a function of x).

But I can't find a relation between them.

My accomplishments

I succeeded in converting the given integral to the following equivalents:

Volume projected onto the Z-X plane ($dz\, dx\, dy$)

This requires the limits of y to be scalars. $$ \int\limits_{0}^{4} \int\limits_{y}^{2y} \int\limits_{\frac{y}{2}}^{2} f_{(x,y,z)}\; dz\, dx\, dy $$ WA's result.

Volume projected onto the X-Z plane ($dx\, dz\, dy$)

$$ \int\limits_{0}^{4} \int\limits_{\frac{y}{2}}^{2} \int\limits_{y}^{2y} f_{(x,y,z)}\; dx\, dz\, dy $$ WA's result.

Volume projected onto the Z-Y plane ($dz\, dy\, dx$)

$$ \int\limits_{0}^{4} \int\limits_{\frac{x}{2}}^{x} \int\limits_{\frac{y}{2}}^{2} f_{(x,y,z)}\; dz\, dy\, dx + \int\limits_{4}^{8} \int\limits_{\frac{x}{2}}^{4} \int\limits_{\frac{y}{2}}^{2} f_{(x,y,z)}\; dz\, dy\, dx $$ WA's result for 1st integral and WA's result for 2nd integral.

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    $\begingroup$ Why are you writing $f_{(x,y,z)}$ instead of $f(x,y,z)$? $\endgroup$
    – zhw.
    May 22, 2015 at 17:28
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    $\begingroup$ @zhw.: Only because it's less important (less important - reduce font size) $\endgroup$
    – Dor
    May 22, 2015 at 17:35
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    $\begingroup$ @Dor No. $f_{(x,y,z)}$ means something entirely different from $f(x,y,z)$. You should stick to common notation, that is either leave out the arguments ($\iiint f$) or write them like all people do. $\endgroup$
    – AlexR
    May 26, 2015 at 18:33
  • $\begingroup$ @AlexR: Ok. I didn't want the post to look "heavy".. Though in most cases people relate by context... $\endgroup$
    – Dor
    May 26, 2015 at 18:49
  • $\begingroup$ @Dor To accomplish that, you could remove all the unnecessary boldfacing and reorder the post so that the relevant part is at the top and the complementary images are at the bottom. You could even leave $f$ out entirely since the question already makes it clear that you do not wish to do any transformation whatsoever. $\endgroup$
    – AlexR
    May 26, 2015 at 18:52

1 Answer 1

Reset to default
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Here we have a 3D picture of the entire region ($x$ horizontal, $y$ depth, $z$ height):

Entire region specified

In blue we see the $x$-$y$ plane, while the pink part stretches both through the $y$-$z$ plane and the $x$-$y$ plane. This side is determined by the equation $x = y$, so the image can be a bit off-putting. Suppose we cut at $y = 2$, then we obtain:

Cut at $y = 2$

Here we see what's going on: for fixed $y$, our 3D shape actually is a square! In other words: given $y$, the dependence of $z$ on $x$ or vice versa is utterly trivial: the bounds do not change with varying $z$ or $x$.

This is of course already implied by the equations we obtained, but it's probably more convincing to see this actually happening. I hope that clears the air for you.

The images are courtesy of Mathematica 8's RegionPlot3D command:

RegionPlot3D[
 y <= x <= 2 y && 0 <= y <= 2 z && 0 <= z <= 2, {x, 0, 8}, {y, 0, 
  8}, {z, 0, 2}, Mesh -> None, PlotPoints -> 200]

RegionPlot3D[
 y <= x <= 2 y && 2 <= y <= 2 z && 0 <= z <= 2, {x, 0, 8}, {y, 0, 
  8}, {z, 0, 2}, Mesh -> None, PlotPoints -> 200]
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    $\begingroup$ Thanks! I appreciate the detailed answer! $\endgroup$
    – Dor
    May 26, 2015 at 20:57

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