Calculate $$\iiint_V \frac{1}{(x+y+z)^3} \, dV$$ Where $V$ is the volume bounded by the planes $$\{4x + 3z = 12, 4x + z = 4, 4y + 3z = 12, 4y + z = 4, z = 0\}$$
I've used simple coordinates change to $z$ such that: $z=4-w$
So the boundris are a little clearer, from here I calculated that the boundaries are: $$0\le w \le4$$
$$ \frac{w}{4} \le x,y\le \frac{3w}{4} $$
with $dz=-dw$
So from here the integral and the boundaries look like this:
$$ \int_0^4 \int_{w/4}^{3w/4} \int_{w/4}^{3w/4}\frac{-dx\,dy\,dw}{(x+y+4-w)^3} $$
But everywhere I tried to even just numerically calcualte this integral I was told the integral is divergent. I know the Boundaries are compact, it's a tilted upside down pyramid but it seems I did a mistake along the way. What do?
The region is a square pyramid with the vertex offset from the base.
i.e. $(1,1,0), (1,3,0), (3,3,0), (3,1,0), (0,0,4)$
We would could solve this by breaking up the integral into regions that are easier to integrate.
If we make a change of coordinates like
$x = u - \frac 12 z\\ y = v - \frac 12 z\\ z = z$
Then our boundaries become
$4u + z = 12\\ 4u -z = 4\\ 4v + z = 12\\ 4v - z = 4\\ z = 0$
Moving our vertices to something centered over the base. $(1,1,0), (1,3,0), (3,3,0), (3,1,0), (2,2,4)$
The integrand: $(x+y+z) = (u+v)$
The jacobain $\det \begin{bmatrix} 1&0&-\frac 12\\0&1&\frac 12\\0&0&1\end{bmatrix} = 1$
$\int_0^4\int_{1+\frac{z}{4}}^{3-\frac {z}{4}}\int_{1+\frac{z}{4}}^{3-\frac {z}{4}} \frac {1}{(u+v)^3}\ du\ dv\ dz$