Volume of tetrahedron using triple integral

Find the volume between $x=0\\y=0\\z=0$ and $2x+y+z=2$ using triple integral

Attempt:

$$\int_0^2\int_0^2 \int _0^{(2-y-z)/2} \, dx \, dy \, dz=\cdots$$

I'm not so sure about choosing the limits for the middle integral, can someone please explain me how should I "see" what should be the limits for the middle one?

• Graph the problem.. – Mattos Aug 4 '16 at 15:11
• What do you mean by the area surrounded by the three planes given? Are you not talking about volume? – zoli Aug 4 '16 at 15:40

1. The outermost integration is on $z$. Therefore you are summing all the horizontal triangular cross-sections of the tetrahedron; each is given by the bounding lines $x=0$, $y=0$, $2x+y=2-z$. The integration limits are, as you noted, $0\leqslant z\leqslant 2$.
2. The next inner integration is on $y$. Therefore you are summing all the linear cross-sections of the $z$-th triangle; each is given by the bounding points $x=0$, $x = (2 - z - y)/2$. The integration limits are $0\leqslant y\leqslant 2 - z$.
3. The innermost integration is on $x$. Therefore you are summing all the point cross-sections of the $y$-th line segment. The integration limits are, as you noted, $0 \leqslant x \leqslant (2 - z - y)/2$.
So the upper bound for the integration on $y$ is $2 - z$.
You have $$\int_0^2 \left( \int_0^\text{what?} \cdots \, dy \right) \,dz$$ With $z$ fixed at some number between $0$ and $2$, the value of $y$ must be between $0$ and $2-z$. If the sum of $y$ and $z$ exceeded $2$, then $2x+y+z$ could not be equal to $2$ unless $x$ were negative, which is not allowed.