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Evaluate $\displaystyle\iiint\limits_E \, \displaystyle\frac{\mathrm{d}V}{(x+y+z)^{3}}$, where E is the region bounded by the six planes $z=1$, $z = 2$, $y = 0$ , $y = z$ , $x = 0$ , $x = y + z$.


I am not sure on how to evaluate this integral. I have calculated a temptative answer, but I'm not sure of it. I assumed that in this case, we were dealing with a Type III region (where $y=f(x,z)$), but I am not sure of the reasoning behind it.

I set up the iterated integral as follows:

$\displaystyle\int_1^2 \displaystyle\int_0^{2z} \displaystyle\int_{z}^{x-z} \! \displaystyle\frac{1}{(x+y+z)^{3}} \, \mathrm{d}y \mathrm{d}x \mathrm{d}z$

Which at the end will yield a result of $I = 2\ln2$. However, the main problem (if it is right), is that I do not understand the reasoning on determining whether a region is a Type I ($z=f(x,y)$, Type II ($x=f(y,z)$) or Type III region. Is there a foolproof way to determine this?

Thank you very much in advance.

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I have limited my graphs to two-dimensional ones, which will serve the purpose of illustrating the type of region is involved in this integration, since this is something you've asked about. (A three-dimensional plot turned out to be somewhat difficult to read.)

For triple integrals, a region is Type I if surfaces between which the integration is to be performed have a common "projection" onto the $ xy-$ plane $ \ ( z = 0 ) \ $. We would describe these surfaces using two functions $ \ z = f_1(x,y) \ $ and $ \ z = f_2(x,y) \ $ and the integration would be set up as $ \ \iint \int_{z_1}^{z_2} \phi(x,y,z) \ \ dz \ (dy \ dx) \ $ , the last two differentials being in either order.

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For the integration in question here, we can look at the domains delimited on the planes $ \ z = 1 \ $ and $ \ z = 2 \ $ and find that their projections onto the $ \ xy-$ plane would differ. So this integration is not being conducted in a Type I region.

Correspondingly, a region is Type II if there is a common projections onto the $ yz-$ plane $ \ ( x = 0 ) \ $ , with the surfaces being two functions $ \ x = f_1(y,z) \ $ and $ \ x = f_2(y,z) \ $ and the integration being $ \ \iint \int_{x_1}^{x_2} \phi(x,y,z) \ \ dx \ (dy \ dz) \ $ . For a Type III region, the common projections are onto the $ xz-$ plane $ \ ( y = 0 ) \ $ , the surfaces are $ \ y = f_1(x,z) \ $ and $ \ y = f_2(x,z) \ $ , and the integration, $ \ \iint \int_{y_1}^{y_2} \phi(x,y,z) \ \ dy \ (dx \ dz) \ $ . [In practice, we don't generally worry about how these regions are labeled, but simply check to see how the integral may be arranged. As you might imagine, integration could occur over a three-dimensional region which falls into none of these categories.]

enter image description here

Returning to our integral again, we can look at the domains where the volume meets the $ \ y = 0 \ $ and $ \ y = 2 \ $ planes to see that we do not have a Type III region. However, the oblique plane $ \ x = y + z \ $ causes the volume to extend as far in the $ \ x-$ direction as $ \ x = 4 \ $ .

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If we "sight down" the $ \ x-$ axis back toward the $ \ yz- \ $ plane, this oblique plane "approaches" us running left to right (toward increasing values of $ \ y \ $) and we find that there is a common projection onto the $ \ x = 0 \ $ plane, making the volume a Type II region. The bounding surfaces are thus the planes $ \ x = 0 \ $ and $ \ x = y + z \ $ . We would then set up and compute the triple integral

$$ \int_1^2 \int_0^z \int_0^{y+z} \ \frac{1}{(x + y + z)^3} \ \ dx \ dy \ dz \ \ = \ \ \int_1^2 \int_0^z \ \left[ \ -\frac{1}{2 \ (x + y + z)^2} \ \right] \vert_{x=0}^{x=y+z} \ \ dy \ dz $$

$$ = \ \ \frac{1}{2} \ \int_1^2 \int_0^z \ \left[ \ \frac{1}{ (0 + y + z)^2} \ - \ \frac{1}{ ([y+z] + y + z)^2} \right] \ \ dy \ dz $$

$$ = \ \ \frac{1}{2} \ \int_1^2 \int_0^z \ \frac{3}{ 4 \ (y + z)^2} \ \ dy \ dz \ \ = \ \ \frac{3}{8} \ \int_1^2 \ \left[ \ -\frac{1}{y + z} \ \right] \vert_{y=0}^{y=z} \ \ dz $$

$$ = \ \ \frac{3}{8} \ \int_1^2 \ \left[ \ \frac{1}{z} \ - \ \frac{1}{2z} \ \right] \ \ dz \ \ = \ \ \frac{3}{8} \ \int_1^2 \ \frac{1}{2z} \ \ dz \ \ = \ \ \frac{3}{16} \ \ln \ z \ \vert_1^2 \ = \ \frac{3}{16} \ln \ 2 \ \ . $$

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