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The function given is

$f(x,y,z) = \displaystyle\frac{1}{(x+y+z+1)^2}$

$D = \{(x,y,z) \in \mathbb{R}^3 : x \geq 0, y \geq 0, z \geq 0, x+y+z \leq 1 \}$

It seems that the domain would be the under a plane that contains the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$; then if i consider the plane $XZ$ i have the set $D_{XZ}= \{(x,z) : z=1-x, 0 \leq x \leq 1 \}$ also $D_{XY}= \{(x,z) : y=1-x, 0 \leq x \leq 1\}$ and $D_{YZ}= \{(x,z) : z=1-y, 0 \leq y \leq 1\}$.

Then i could set $0\leq x\leq z-1$, $0\leq y\leq z-1$ and $0\leq z\leq 1$, and the integral would be given by:

$$\begin{align*} \displaystyle\int\displaystyle\int\displaystyle\int_D f(x,y,z)\,dx\,dy\,dz &= \displaystyle\int\displaystyle\int\displaystyle\int_D \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz \\&= \displaystyle\int_0^1\displaystyle\int_0^{1-z}\displaystyle\int_0^{1-z} \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz \\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{(1-z)+y+z+1)}- \displaystyle\frac{1}{(0)+y+z+1)}\bigg)\,dy\,dz \\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{y+2}- \displaystyle\frac{1}{y+z+1}\bigg)\,dy\,dz \\ &= -\displaystyle\int_0^1 (\log(y+2)-\log(y+z+1))|_{y=0}^{y=1-z}\,dz \\ &= -\displaystyle\int_0^1 \{[(\log(3-z)-\log(2)] - [\log(2)-\log(z+1)]\}\,dz \\ &= -\displaystyle\int_0^1 [(\log(3-z)+\log(z+1)-2\log(2)]\,dz \\ &= 2\log(2)+\{[(3-z)\log(3-z) \\ & \;\;\;- (3-z)]|_{z=0}^{z=1} - [(z+1)\log(z+1)-(z+1)]|_{z=0}^{z=1}\} \\ &= 2\log(2) - 3\log(3) +4 \end{align*} $$

Did i solved it correctly?

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1 Answer 1

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I don't think so. I think the integral looks like

$$\int_0^1 dx \, \int_0^{1-x} dy \, \int_0^{1-x-y} dz \frac{1}{(1+x+y+z)^2}$$

which is equal to

$$-\int_0^1 dx \, \int_0^{1-x} dy \, \left [\frac{1}{2} - \frac{1}{1+x+y} \right ]$$

which is equal to

$$-\int_0^1 dx \, \left [\frac12 (1-x) - \log{2} + \log{(1+x)} \right ] $$

which in turn is equal to

$$-\frac14 + \log{2} - \left [ (1+x) \log{(1+x)} - (1+x)\right]_0^1 = \frac{3}{4}-\log{2}$$

This result was verified in Mathematica.

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