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
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, $I = -3 V = -18$