Calculating $ \int_{E}(y-3) \, dx \, dy \, dz$ So I have the set $E$ as the smallest polyhedron that contains the points $(-1,0,0), (0,2,0), (0,-2,0), (2,0,0)$ and $(0,0,3)$. And I have to calculate $$ \int_{E}(y-3) \, dx \, dy \, dz$$ In all of these types of exercises is for me an extremely hard thing to find the borders of $x$, $y$ and $z$.
The set $E$ according to what I get would be a pyramid, and I got the borders directly from the given points and the way how this pyramid looks when I draw it and I calculated the Integral:
$$ \int_{0}^{3}\int_{-2}^{2}\int_{-1}^{2}(y-3) \, dx \, dy \, dz =-108$$
The right answer according to my book would be $-18$. What am I doing wrong? What is the problem with my borders and what would be some tips to be able to calculate them in these types of exercises?
Thanks in advance
 A: Given set $E$: If we take the base of the pyramid in $XY$ plane, it is formed by lines joining the points $A(-1, 0, 0), B(0,2,0), C(2,0,0)$ and $D(0,-2,0)$. The vertex of the pyramid is $P(0,0,3)$.
In $XY$ plane this translates to,
$AB: y-2x = 2, BC: x+y = 2, CD: x-y=2, DA: 2x+y = -2$
Given the vertex $P$ of the pyramid, the equation of the plane that face $ABP$ lies on is $y-2x+\frac{2}{3}z = 2$
Equation of the plane that face $BCP$ lies on is $x+y+\frac{2}{3}z = 2$
We will not need the other two planes below $x-$axis.
Now coming to the integral we need to work upon is $\displaystyle \iiint_E (y-3) \ dv$
But if you look at the given set $E$, it is completely symmetric to $y-$plane ($XZ$ plane) and given $y$ is an odd function, integral of $y$ over the volume of set $E$ will be zero. So our integral reduces to,
$I = -3 \displaystyle \iiint_E  dv$, in other words if we just find the volume of the pyramid, we are done which is,
$V =  2 \displaystyle \int_0^3 \int_{0}^{(6-2z)/3} \frac{6 - 3y - 2z}{2} \ dy \ dz = 6$
(we integrate over the region above $y-$ plane and multiply the volume by $2$ as the set is symmetric to $y-$plane as I mentioned earlier.
So, $I = -3 V = -18$
