The region in $\mathbb{R}^3$ bounded by the planes $y = 1$, $y = −x$, $z = −x$ and, the coordinate planes, $x = 0$ and $z = 0$
$$ \int_{0}^{1} \int_{0}^{-y}\int_{0}^{-x} \;dz \;dx\;dy $$ I feel like I am a bit off with the bounds here?
Can someone please confirm?
On
bounded by the planes $y = 1, y = −x, z = −x$ and, the coordinate planes, $x = 0$ and $z = 0$
Well, if we talk about the $(x,y)$-ordinates alone for the moment, the lines $y=1, y=-x, x=0$ form the boundaries of the right triangle: $\triangle(-1,1)(0,1)(0,0)$ which is $\{(x,y):x\in(-1;0), y\in(-x;1)\}$
Now, project this triangle up above the plane $z=0$ and below the diagonal plane $z=-x$. That is: $\{(x,y,z):x\in(-1;0), y\in(-x;1), z\in(0;-x)\}$, which is the corner slice of a cube.
There you go.
$\displaystyle\color{floralwhite}{\int_{-1}^0\int_{-x}^1\int_0^{-x} 1\operatorname d z\operatorname d y\operatorname d x = \tfrac 1 6}$
Here's a plot of the bounded region:
Your integral setup isn't correct (at least not in the current order of your variables - I haven't checked). It should be $$\int_0^1 \int_{-y}^0\int_0^{-x}\mathrm{d}z\,\mathrm{d}x\,\mathrm{d}y$$