What is the measure of $A=[-1,2]\times[0,3]\times[-2,4]\cup[0,2]\times[1,4]\times[-1,4] \setminus [-1,1]^3$? I really get stuck after one point, and don't know where to go on.I know that my try, up to where I am stuck is correct.
$$\color{#20f}{\text{TRY:}}$$ $$B_1=[-1,2]\times[0,3]\times[-2,4],\mu(B_1)=3 \cdot 3\cdot6=54 \\
              B_2=[0,2]\times[1,4]\times[-1,4] \mu(B_1)=2 \cdot 3\cdot5=30$$
$\mu(A)=\mu(B_1\cup B_2)=\mu(B_1)+\mu(B_2)-\mu(B_1\cap B_2).$
and $B_1\cap B_2=([-1,2]\cap[0,2])\times([0,3]\cap[1,4])\times([-2,4]\cap[-1,4])=[0,2]\times [1,3] \times[-1,4]$
$$\mu (B_1\cap B_2)=2 \cdot 2\cdot5=20$$
therefore: $$\color{#f00}{\mu(A)=54+30-20=64.}$$
Now that I know that this method is correct how do I find the measure of $$B=A\setminus [-1,1]^3$$ I thought $\mu(B)=\mu(A)-\mu (A \cap [-1,1]^3)$ yet I have no clue what $A \cap [-1,1]^3$ is. Help?
 A: Method two is wrong. In general, you don't have $(A_1 \times B_1) \cup (A_2 \times B_2) = (A_1 \cup A_2) \times (B_1 \cup B_2)$ since
$$ (A_1 \cup A_2) \times (B_1 \cup B_2) = (A_1 \times B_1) \cup (A_1 \times B_2) \cup (A_2 \times B_1) \cup (A_2 \times B_2) $$
and the right hand side has two extra possibly non-empty terms.
Method one works since indeed you have $(A_1 \times B_1) \cap (A_2 \times B_2) = (A_1 \cap A_2) \times (B_1 \cap B_2)$.
A: Let
$$A:=[-1,2]\times[0,3]\times[-2,4],\quad B:=[0,2]\times[1,4]\times[-1,4],\quad C:=[-1,1]^3\ .$$
I shall compute the volume of $(A\cup B)\setminus C$, which is not the same as $A\cup(B\setminus C)$.
The projection of $A\cup B$ onto the $(x,y)$-plane is the union of the two rectangles $A':=[-1,2]\times[0,3]$ and $B':=[0,2]\times[1,4]$ of areas $9$ and $6$ respectively. Since $A'\cap B'=[0,2]\times[1,3]$ has area $4$ it follows that the area of $P':=A'\cup B'$ is $9+6-4=11$. Now $A$ has its base at level $z=-2$, $B$ at $z=-1$, while both have their top at $z=4$. It follows that
$${\rm vol}(A\cup B)={\rm area}(P')\cdot 5+{\rm area}(A')\cdot 1=64\ .$$
Now we have to subtract the cube $C$. Since the part $y<0$ of $C$ is outside of $A\cup B$ and the part $y\geq0$ of $C$ is contained in $A$ we have to subtract ${1\over2}{\rm vol}(C)=4$ from our intermediate result. In this way the volume of the set in question is seen to be $60$.
