Find the volume of the region in the square column $|x| + |y| \leq 1 $ and bounded by $3x+z=3, z=0$
I have attempted to find the bounds, by letting $z=3-3x, \text{ then letting } z=0, \text{ to give } 3x=3 \text{ and } x=1$ as an upper bound for x
Then I arrived at the integral:$\int_{x=-1}^1\int_{y=1-|x|}^1\int_{z=0}^{3-3x}dzdydx$
I then evaluated this to be $3$.
This doesn't feel correct, I feel like I may have found the bounds incorrectly, it didn't feel correct having $y=1-|x|$ as the lower bound.
Since $|x| +|y| =1 $ is the boundary then
$ - (1 - |x| ) \le y \le 1 - |x| $
where $-1 \le x \le 1 $
The volume integral is
$V =\displaystyle \int_{x=-1}^{1} \int_{y = -(1 - |x|)}^{1 - |x|} \int_{z=0}^{z = 3 - 3 x } dz dy dx $
This simplifies to,
$V =\displaystyle \int_{x=-1}^{1} \int_{y = -(1 - |x|)}^{1 - |x|} ( 3 - 3 x ) dy dx $
Integrating with respect to $y$ yields,
$V =\displaystyle \int_{x=-1}^{1} \int_{y = -(1 - |x|)}^{1 - |x|} (3 - 3x)(2)(1 - |x|) dx $
Finally integrating with respect to $x$ and noting that $x |x| $ is odd, this becomes,
$V = 6 \displaystyle \int_{-1}^{1} ( 1 - |x| - x (1 - |x| ) ) dx = 6 ( 2 - 1 - 0 + 0) = 6 $
Another way to compute the volume is as follows:
The midpoint of $z$ is at $(0,0)$ and is equal to $3$ , so the volume is $3 \times (\sqrt{2})^2 = 6 $