so my textbook gave the following exercise:
Calculate the volume of the area which is bounded by the following surfaces: $x=1$, $x=3$, $y=1$, $y=5$, $z=0$, $z=4-2x$
As a hint, they told us that we had to divide the section into two parts, the first one where $z>0$ and the second one where $z<0$.
This is my approach so please tell me if I am correct or incorrect, and I'd appreciate if someone would provide help if I am incorrect
I evaluated the following integrals:
$$ \int_{1}^{5} \int_{1}^{3} |4-2x|dxdy$$
$|4-2x| = 4-2x$ for $x \le 2$
$|4-2x| = 2x-4$ for $x > 2$
Which amounts to:
$$ \int_{1}^{5} \int_{1}^{2}(4-2x)dxdy + \int_{1}^{5} \int_{2}^{3} (2x-4)dxdy$$
Both integrals evaluate to 4, so the sum is $8$.
Is this correct?