I'm working through some practice problems on volume integrals while preparing for an exam, and I don't understand the bounds in one of them. Here's the prompt in the textbook:
Find the volume of the region bounded by $z=x+y, 0\leq x\leq 7, 0 \leq y\leq 7$, and the planes $x=0, y=0$, and $z=0$.
I know (from the solutions manual and my professor) that the answer is given by $\int_0^7\int_0^{7-x}\int_0^{x+y} dzdydx$ (and they don't explain why), but why is $7-x$ the upper bound for $y$? I thought it would be just $7$.
Edit: I am aware that the bounds for x and y on this integral give us a triangle in the first quadrant of the xy-plane with the axes and the line y=7-x as its sides, so maybe I should have been more specific with my question:
How could my professor and the solutions manual be interpreting the instructions (especially the inequalities) to get the triangle and not the square on the xy-plane?
For $[0,7]\times [0,7]$ in $OXY$ plane with roof $z=x+y$, floor $z=0$ volume is $$\int\limits_0^7\int\limits_0^{7}\int\limits_0^{x+y} dzdydx$$ Addition 1. this integral corresponds exactly to $$\left\{\begin{array}{} 0 \leqslant x \leqslant 7 \\ 0 \leqslant y \leqslant 7 \\ 0 \leqslant z \leqslant x+y \end{array}\right\}$$ which is in OP and is exactly answer to OP and is volume over whole rectangle, not only it's lower triangle .
Addition 2. If we want to have volume over triangle, then it is set: $$\left\{\begin{array}{} 0 \leqslant x \leqslant 7 \\ 0 \leqslant y \leqslant 7-x \\ 0 \leqslant z \leqslant x+y \end{array}\right\}$$ so integral will be $$\int\limits_0^7\int\limits_0^{7-x}\int\limits_0^{x+y} dzdydx$$ excuse me for repeating, that it is not initial OP.