Lebesgue measure of a 3-Simplex

Question

I try to calculate the Lebesgue measure of this set: $ S_3:=\{(x,y,z)\in \mathbb{R}^3:x+y+z\leq 1\} $. Here is a picture of the set:

$ S_3 $" />

At fisrt I set the integration limits:

$ x\leq 1-y-z $ so $ 0\leq x \leq 1-y-z $

$ y\leq 1-x-z $ so $ 0\leq y \leq 1-x-z $

and $ 0\leq z \leq 1 $

As far correct?

Then I integrate three times:

$ \begin{align} \lambda^3(S_3)&=\int\limits_{S_3}1\ d\lambda^3\\&=\int\limits_0^1\Bigg(\int\limits_0^{1-x-z}\Bigg(\int\limits_0^{1-y-z} 1\ dx\Bigg)dy\Bigg)dz \\&=\int\limits_0^1\Bigg(\int\limits_0^{1-x-z}\Bigg(1-y-z\Bigg)dy\Bigg)dz\\&=\int\limits_0^1\Bigg(\dfrac{\left(z-x-1\right)\left(z+x-1\right)}{2} \Bigg)dz\\&=-\dfrac{3x^2-1}{6} \end{align} $

But I has to be $ \lambda^3(S_3)=\frac{1}{6} $. What went wrong here?

Answer 1

Your integration limits should be $$\int_0^1 \left( \int_0^{1-z} \left( \int_0^{1-y-z} 1 \, dx \right) dy \right) dz.$$ You have an extra $x$ lying around in the integration limit.

Edit to answer comment: When you are performing a multiple integration you are integrating one variable while leaving other constants. In this problem you performed it in the following sequence: $$\begin{aligned} g(y,z) & = \int_m^n f(x,y,z) \, dx \\ h(z) & = \int_c^d g(y,z) \, dy \\ V & = \int_a^b h(z) \, dz \end{aligned}$$

In this particular case, you are integrating slices of a plane perpendicular to the $x$-axis up to the tetrahedron face where $x+y+z=1$. Once you are done with this integration $x$ no longer plays a role. Next, you are integrating on the $yz$ plane from the $y$ axis to the line where $y+z=1$. Finally, you find that value for $z$ in the interval $[0,1]$.

Answer 2

You seem a little turned around in your ranges of the variables. Since, in this problem, the three variables are interchangeable, let us assume that $z$ is the last (outermost integral) variable, $y$ is the second (second integral) variable and $x$ is the first (innermost integral) variable.

This means that the bounds on the $z$ integral cannot depend on any other variables. By inspecting the diagram, we see $0 \leq z \leq 1$. $$ \int_0^1 \cdots \,\mathrm{d}z $$

It means the bounds on the $y$ integral can only depend on $z$. The inner two integrals are evaluated at a fixed value of $z$, so pick a slice of constant $z$ in the diagram. In that slice, the least value of $y$ is $0$ and the greatest is $1-z$. $$ \int_0^1 \left( \int_0^{1-z} \cdots \,\mathrm{d}y \right) \,\mathrm{d}z $$ (Outside the parenthesized expression, the dummy variable $y$ has no value. Inside the parenthesized expression, $z$ has a value. The bounds of the $y$-integral can depend on $z$, but not vice versa.)

It means the bounds on the $x$ integral can depend on $z$ and $y$. Return to the diagram. On the slice of fixed $z$, also fix $y$, yielding a line segment parallel to the $x$-axis with $x$-coordinates running from $0$ to $1 - y - z$. $$ \int_0^1 \left( \int_0^{1-z} \left[ \int_0^{1-z-y} 1 \,\mathrm{d}x \right]\,\mathrm{d}y \right) \,\mathrm{d}z $$ ($x$ and $y$ don't exist outside the parentheses. Inside the parentheses, $z$ exists, so can be used in the $x$- and $y$- integrals. $x$ doesn't exist outside the brackets. Inside the brackets, $z$ and $y$ exist so both can appear in the bounds on $x$.)

One normally does not put the parentheses and brackets in, so $$ \int_0^1 \int_0^{1-z} \int_0^{1-z-y} 1 \,\mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z $$