Write out the triple integral for $f(x,y,z)$ to describe a region bounded by the planes $x=0, \, y=0,\, z=0, \, x+y=4$, and $x=z-y-1$.
I plugged in the 4th plane into the 5th plane and solved $z=5$. So, I know z is between 0 and 5 so it is the outermost integral.
I'm not entirely sure how to write out the bounds for x and y. Can someone help me with this?
The lines $x = 0$, $y = 0$ and $x + y = 4$ form a right-angled triangle with vertices $(0,0), (4,0)$ and $(0,4)$ on the $xy$-plane. Therefore, $0 \le x \le 4$ and $0 \le y \le 4 - x$. Make $z$ the subject in the equation for the 5th plane: $z = x + y + 1$. Denote the region as $R$
$$\therefore \iiint_R f(x,y,z) \,dx dy dz = \int_0^4 \int_0^{4-x} \int_0^{x+y+1} f(x,y,z) \, dz dy dx$$