Volume of the region bounded by $z=1-x-y$

Question

This is a very standard exercise in Calculus. The point of this question is, well, I'm able to solve it, but I'm struggling to make out why I'm choosing the integration ranges the way they are chosen. A lot of similar exercises are asked (e.g. volume of the region bounded by the surfaces $x = 0, y = x, y = 2 - x^2, z = 0$ and $z = x^2$, Volume of the first octant under a surface), but I'm trying to understand this specific problem.

Find the volume of the region in the first octant bounded above by the surface $z=1-x-y$.

Obvious Solution: $$ V=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} 1 d z d y d x=\int_{0}^{1} \int_{0}^{1-x}(1-x-y) d y d x=\int_{0}^{1} \frac{(1-x)^{2}}{2} d x=\frac{1}{3} . $$ My thinking for choosing the integration ranges like that isn't clear. I've visualized and drawn what that volume looks like. As it seemed like $z$ increased from $0$, and $z = 1-x-y$, I chose $\int_{0}^{1-x-y} 1 d z$. (This probably isn't a good way to think about things) But now I get confused for integrating over $y$. I get that $0 \le x \le 1$, $0 \le y \le 1-x$, $0\le z \le 1-x-y$ describes $V$ if I draw it in Matlab or something, but how can I justify this for certain using just reasoning, and a pencil and paper? I should note that I'm expected to be able to do this quickly under exam conditions.

Answer 1

Those ranges are determined from the exterior to the interior, not the other way around.

So, let us begin with $x$. Your region is the tetrahedron $T$ whose vertices are $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. If $(x,y,z)\in T$, then $x\in[0,1]$. So, your integral is of the form$$\int_0^1(\ldots)\,\mathrm dx.$$Now, let us consider $y$. For each $x$, you have $y\geqslant0$ and $y+z\leqslant1-x$. In particular (since $z\geqslant0$), $y\leqslant1-x$. It turns out that $(x,y,z)\in T\iff y\in[0,1-x]$. So, your integral is of the form$$\int_0^1\int_0^{1-x}(\ldots)\,\mathrm dy\,\mathrm dx.$$Finally, for each $x$ and each $y$, $z\geqslant0$ and $z\leqslant1-x-y$. It turns out that $(x,y,z)\in T\iff y\in[0,1-x-y]$. So, your integral is of the form$$\int_0^1\int_0^{1-x}\int_0^{1-x-y}(\ldots)\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$

Answer 2

Think of it like this. a cylinder can be made up by stacking coins of radius upto a particular height. This is the kind of thinking we want to use , but instead of circles(coins), we want to stack the areas which is formed by the projection of the the face of the trapezoid in the $xy$ plane. So in general if $z=f(x,y)$ then you first define the "height" upto which you want to stack(the height is a function of x,y). So the limits for $z$ is $0\leq z\leq f(x,y)$ . Then you want to find the area projection of the surface $z=f(x,y)$ , and then stack those areas together. The projection is given by just $f(x,y)=0$ which represents a 2D curve. Now you just use double integral of $f(x,y)$, over the region bound by the curve $f(x,y)=0$ . So in this case your $f(x,y)=1-x-y$. And for the area bounded $1-x-y=0$ , you get the following two limits any of which is valid :- $\int_{y=0}^{1}\int_{x=0}^{1-y}(1-x-y)dxdy$ or $\int_{x=0}^{1}\int_{y=0}^{1-x}(1-x-y)dydx$.

You can generalize even further :- for the volume bounded by surfaces $z=f(x,y)$ and $z=g(x,y)$ ,i.e. ($g(x,y)\leq z\leq f(x,y)$) You get :-

$\iint_{S}\int_{g(x,y)}^{f(x,y)}dzdxdy$ where $S$ denotes the area bounded by the curve of intersection $f(x,y)=g(x,y)$ or $f(x,y)-g(x,y)=0$ .

Hope this helps

Answer 3

One of the ways to think about it is to look at the projection in xy-plane.

The surface is $x + y + z = 1$ and $x, y, z \geq 0$

Rewriting, $x + y = 1 - z$ and given $x, y, z$ are non-negative, $x + y$ is maximum in xy-plane (when $z = 0)$. Now let's focus on the projection in xy-plane (see the diagram).

For any point in xy-plane within the shaded region, think of a vertical strip parallel to z-axis. It is bound between $z = 0$ and plane $x + y + z = 1$. So for every point in the shaded region, $0 \leq z \leq 1-x-y$.

Now for the area of the shaded region, $0 \leq y \leq 1 -x, 0 \leq x \leq 1$ or $0 \leq x \leq 1 -y, 0 \leq y \leq 1$, depending on the order of integral we choose. So one of the ways we can write the integral to find volume is,

$\displaystyle V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} \ dz \ dy \ dx$