Let $b$ a real positive number. Calculate the volume of the tetrahedrum delimited by the planes $x=0, y=0, z=0$ and $2bx+2y+z=2b$.
The first thing i've done is calculating the vertices of which are $(0,0,2b), (0,b,0), (1,0,0)$. Then, I want to see how the variables move.
First, we know $0 \leq x \leq 1$. Now, looking the $xy$ plane, we see that $0 \leq y \leq b-bx$.
The problem is with $z$, i found 2 ways to move it but the final integral ends up giving me different results. So, the first way is isolating $z$ from $2bx+2y+z=2b$, where we get $z = -2bx-2y+2b$. Then, the volume is
\begin{equation} \int\limits_{0}^{1} \int\limits_{0}^{b-bx} \int\limits_{0}^{-2bx-2y+2b}1 \,dz\, dy\,dx = \frac{b^2}{3} \end{equation}
On the other hand, if we look at the zy plane, we see that $z = 2b-2y$. In this case, we have
\begin{equation} \int\limits_{0}^{1} \int\limits_{0}^{b-bx} \int\limits_{0}^{2b-2y}1 \,dz\,dy\,dx = \frac{2b^2}{3} \end{equation}
What is the correct way to calculate it? Where is my mistake?