The volume of two vertically stacked, diagonally halved unit cubes

Question

How do I express the volume of the painted area using multiple integrals? The answer should be $1$, because the volume is $0.5 \times 2 \times 1 \times 1=1$.

Answer 1

It is obvious that $0\leq z \leq 2$.

Let $f(t)$ be the area of the intersection of red area and $z=t$.

Then the volume is $\int_0^2 f(t) dt$.

the intersection of red area and $z=t$ is expressed as follows:

$$1- \frac{1}{2}t \leq x \leq 1, ~~ 0 \leq y \leq 1.$$

Thus $f(t)$ is

$$ \int_0^1 \int_{1- \frac{1}{2}t}^1 dx dy.$$

Hence, the volume is

$$\int_0^2 \int_0^1 \int_{1- \frac{1}{2}t}^1 dx dy dt = \int_0^2 \int_0^1 \int_{1- \frac{1}{2}z}^1 dx dy dz. $$

Answer 2

You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]\times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume $${\rm vol}(B)=\int_Q f(x,y)\>{\rm d}(x,y)=\int_0^1\int_0^1 2x \>dy\>dx=\int_0^1 2x\>dx=x^2\biggr|_0^1=1\ .$$