triple integral over region defined by the pyramid$ (0,0,0), (1,0,0), (0,2,0), (2,2,0), (1,0,2)$

Question

$∭ z dV(x,y,z)$ over the region defined by the pyramid with base $(0,0,0), (1,0,0), (0,2,0), (2,2,0)$ and vertex $(1,0,2)$.

I tried finding the planes connecting $(0,0,0), (0,2,0), (1,0,2)$ and $(0,2,0), (2,2,0), (1,0,2)$. I found the first plane to be $z = 2x$ and the second to be $z = y+2$.

I then summed the integral from $0$ to $1$ for $x, 0$ to $-2x+2$ for $y, 0$ to $2x$ for $z$ and the integral from 0 to 1 for $x, -2x+2$ to $2$ for $y, 0$ to $y+2$ for z. This turned out to be 6 but the answer is 1 (answer key says to integrate from $0$ to $2$ for $y$, $0$ to $2-y$ for z, and $z/2$ to $1 + y/2$ for x).

Answer 1

A simple solution may go as follows. Let us first focus on the base of the pyramid. It has the following trapezoidal shape:

(0,2)         (2,2)
    *--------*
    |       /
    |      /
    |     /
    *----*
(0,0)    (1,0)

(Sorry for the poor quality, this is only a quick piece of ascii art to pass quickly to the integral.) Its area is $\frac 12(1+2)\cdot2 = 3$.

Now consider all points above enriched with a third component, "$z$", which is zero, and from the vertex $(1,0)$, in the mean time $(1,0,0)$ we draw a perpendicular on the plane $z=0$, and a segment from this point to $(1,0,2)$.

Imagine now $z$ moving linearly from $0$ (base level of the pyramid) to $2$ (height of the vertex which is the corner of the pyramid). At a given level $z$ we have a section $P_z$ of the pyramid, also trapezoidal, a similitude factor w.r.t. the baze $P_0$ equal to $$ \frac 12(2-z)$$ for the lengths, (it is one for $z=0$, zero for $z=2$, and linear in between,) thus a factor $\frac 14(2-z)^2$ for the areas. So we can compute the integral: $$ \begin{aligned} \iiint_P z\; dx\; dy\; dz &= \int_0^2 z\; dz\iint_{P_z}dx\; dy \\ &= \int_0^2 z\; dz\cdot\operatorname{Area}(P_z) = \int_0^2 z\cdot \frac 14(2-z)^2\; dz\cdot\operatorname{Area}(P_0) \\ &= \frac 13\cdot 3 = 1\ . \end{aligned} $$