Triple Integral - Finding the Volume

Question

I'm taking calculus $3$ and already calculated simple volumes with triple integrals. However, I can't do this one. I've gathered that $-1 < z < 1$ but don't know what to do next. Usually I draw a plane after that (for example in cases where $x$ and $y$ don't depend on $z$), but in this case I don't know which plane to choose. Here is the exercise:

Write an expression for the volume of X with the following form: $\int\left(\int\left(\int d x\right) d y\right) d z$.

$X=\left\{(x, y, z) \in \mathbb{R}^{3} : x+y+z<1 ; 0<x<1 ; 0<y<1 ; z>-1\right\}$

Solution:

$\begin{aligned} \operatorname{vol}_{3}(X)=\int_{-1}^{0}\left(\int_{0}^{-z}\left(\int_{0}^{1} d x\right) d y\right) d z &+\int_{-1}^{0}\left(\int_{-z}^{1}\left(\int_{0}^{1-z-y} d x\right) d y\right) d z+\\ &+\int_{0}^{1}\left(\int_{0}^{1-z}\left(\int_{0}^{1-z-y} d x\right) d z\right) d z \end{aligned}$

Thanks for your time.

Answer 1

This solution tries to solve (blindly) the problem using the strategy of isolating inequations in less and less variables, that must have at least one solution. Our start is the set of inequations: $$ \left\{ \begin{aligned} &0<x<1\\ &0<y<1\\ &-1<z<1-x-y\ . \end{aligned} \right. $$ The OP wants us to have a Fubini splitting which depends first on $z$.

OK, which are the values that $z$ may take? The given lower bound remains, for the upper bound we note $z<1-x-y<1-0-0=1$. So $z$ runs in the interval $[-1,1]$.

Now fix such a value of $z$. Which is the portion from the square $(0,1)^2$ where the point $(x,y)$ has to live in, cut by the inequality $x+y<1-z$?

Obviously, this is an "antidiagonal" of the square which passes through a point of the diagonal $x=y$ between $(0,0)$ (for the maximal value of $z$) and $(1,1)$ for the minimal value of $z$. The shape of the region after the cut is

• either a triangle with $(0,0)$ as a vertex of its closure, this happens for $z\in[0,1)$, and this corresponds to the last summand $$ \int_{0}^{1}\left(\int_{0}^{1-z}\left(\int_{0}^{1-z-y} d x\right) d z\right) d z $$ from the original post
• or else, for $z\in(-1,0]$, a region of the shape "square without triangle with the $(1,1)$ in its closure". We have to split it in a cartesian fashion. One piece of it corresponds to the rectangle part, where $x$ fully covers the segment $(0,1)$, this is $$ \int_{-1}^{0}\left(\int_{0}^{-z}\left(\int_{0}^{1} d x\right) d y\right) d z \ , $$ and else a trapezoidal part corresponding to $$ \int_{-1}^0\left(\int_{-z}^1\left(\int_{0}^{1-z-y} d x\right) d y\right) dz\ . $$ Later added picture:

Its area is clear (area of square with side one minus half area of square...), but this is not our problem, our problem is to understand how can we get closer to way an other person was splitting it in parts, so that it is harder to understand what happens.

We fix a value of $z\in(-1,0]$. The antidiagonal $x+y=1-z$ intersects the $(x,y)$-square $(0,1)\times (0,1)$ either in the segments segments that come together in $(0,0)$, or in those that come together in $(1,1)$. The latter case is the more complicated case, and this case is the case for $z\in(-1,0]$. The two sides are cut in $(1,-z)$ and $(-z,1)$.

Consider first the brown rectangle with vertices $(0,0)$, $(1,0)$, $(1,-z)$, $(0,-z)$. Its area is clear, but this is not our problem, our problem is to understand where it appears in the integrals. Here is the place: $$ \left(\int_{0}^{-z}\left(\int_{0}^{1} d x\right) d y\right)\ . $$ Now we pass to the trapezoidal blue part. The $y$-variable goes from $-z$ to one. For each fixed value of $y$, the $x$ variable covers a segment, always starting with $x=0$. The end is on $x+y=1-z$, so we have the other end in $x=1-y-z$. This explains the following part in the other integral: $$ \left(\int_{-z}^1\left(\int_{0}^{1-z-y} d x\right)d y\right)\ . $$

Answer 2

The set $X$ can be written in this form: $$X=\bigl\{(x,y,z)\bigm|0<x<1, \ 0<y<1,\ -1<z<1-x-y\bigr\}\ .$$ While $(x,y)\in\>]0,1[^2$ the admissible $z$ cover an interval of length $$h(x,y)=1-x-y-(-1)=2-x-y\geq0\ .$$It follows that $${\rm vol}(X)=\int_{]0,1[^2}h(x,y)\>{\rm d}(x,y)=\int_0^1\int_0^1 (2-x-y)dy\>dx=1\ .$$