Computing a three-dimensional Lebesgue measure of a bounded set

How can I compute the three-dimensional Lebesgue-measure of the set $A$ which is bounded by the areas $x+y+z =6$, $x=0$, $z=0$ and $x+2y=4$?

A hint on how I solve problems like this in general would be much appreciated.

I need to know how to compute the measure/volume by exclusively using integrals.

• Sorry of being (too?) late, but I've posted another answer that I've been composing anyway. – Han de Bruijn Jun 5 '16 at 18:07

Your set $A$ is a three-sided pyramid standing on the $(x,y)$-plane, and having its top at $(0,0,6)$. The base $B$ of this pyramid is a triangle bounded by the lines $x=0$, $x+y=6$, and $x+2y=4$. Its vertices are $(0,2)$, $(0,6)$, and $(8,-2)$. The set $B$ can be described as follows: $$B:=\left\{(x,y)\>\biggm|\>0\leq x\leq 8, \ 2-{x\over2}\leq y\leq 6-x\right\}\ .$$ The walls $x=0$ and $y=2-{x\over2}$ of $A$ are vertical, whereas the third wall is given by $z=6-x-y=:h(x,y)$. It follows that $${\rm vol}(A)=\int_B h(x,y)\>{\rm d}(x,y)=\int_0^8\int_{2-x/2}^{6-x}(6-x-y)\>dy\>dx=\ldots={64\over3}\ .$$

Isn't that measure the same as the volume of the tetrahedral pyramid, bounded by those areas? The lines $x+y+0=6$ and $x+4y=0$ intersect at $(8,-2,0)$, which is thus one of the vertices of the pyramid. The other vertices are $(0,0,0)$ and $(0,6,0)$ and $(0,0,6)$. The required volume can be calculated in several ways. One way is the area of the triangle at the bottom times one-third of the pyramid's height: $8 \times 6/2 \times 6/3 = \large 48$ .

• Oh wow thanks for the effort, your drawings helped me a lot. I made a mistake though, it should be $x+2y=4$ instead of $x+4y=0$, but that I should be able to fix myself now. Now if I want to compute the volume by using integrals, how do I determine the upper and lower limits of the integrals? – Tesla Jun 3 '16 at 17:10
• And yes, of course it is the same as the volume. But our upcoming exam will be about computing volumes/measures by using integrals (Tonelli etc).. – Tesla Jun 3 '16 at 18:37

AfterMath. By "exclusively using integrals" has been done by Christian Blatter, so here comes the "like this in general" part, as formulated in the question too. Hope it is appreciated by the OP. Linear Tetrahedron

Let's consider the simplest non-trivial finite element shape in 3-D, which is a linear tetrahedron. Function behaviour is approximated inside such a tetrahedron by a linear interpolation between the function values at the vertices, also called nodal points. Let $T$ be such a function, and $x,y,z$ coordinates, then: $$T = A.x + B.y + C.z + D$$ Where the constants A, B, C, D are yet to be determined. Substitute $x=x_k$ , $y=y_k$ , $z=z_k$ with $k=0,1,2,3$. Start with: $$T_0 = A.x_0 + B.y_0 + C.z_0 + D$$ Clearly, the first of these equations can already be used to eliminate the constant $D$, once and forever: $$T - T_0 = A.(x - x_0) + B.(y - y_0) + C.(z - z_0)$$ Then the constants $A$ , $B$ , $C$ are determined by: $$\begin{array}{ll} T_1 - T_0 = A.(x_1 - x_0) + B.(y_1 - y_0) + C.(z_1 - z_0) \\ T_2 - T_0 = A.(x_2 - x_0) + B.(y_2 - y_0) + C.(z_2 - z_0) \\ T_3 - T_0 = A.(x_3 - x_0) + B.(y_3 - y_0) + C.(z_3 - z_0) \end{array}$$ Three equations with three unknowns. A solution can be found: $$\left[ \begin{array}{c} A \\ B \\ C \end{array} \right] = \left[ \begin{array}{ccc} x_1-x_0 & y_1-y_0 & z_1-z_0 \\ x_2-x_0 & y_2-y_0 & z_2-z_0 \\ x_3-x_0 & y_3-y_0 & z_3-z_0 \end{array} \right]^{-1} \left[ \begin{array}{c} T_1-T_0 \\ T_2-T_0 \\ T_3-T_0 \end{array} \right]$$ It is concluded that $A,B,C$ and hence $(T-T_0)$ must be a linear expression in the $(T_k-T_0)$: $$T - T_0 = \xi.(T_1 - T_0) + \eta.(T_2 - T_0) + \zeta.(T_3 - T_0)$$ $$= \left[ \begin{array}{ccc} \xi & \eta & \zeta \end{array} \right] \left[ \begin{array}{c} T_1-T_0 \\ T_2-T_0 \\ T_3-T_0 \end{array} \right]$$ See above: $$= \left[ \begin{array}{ccc} \xi & \eta & \zeta \end{array} \right] \left[ \begin{array}{ccc} x_1-x_0 & y_1-y_0 & z_1-z_0 \\ x_2-x_0 & y_2-y_0 & z_2-z_0 \\ x_3-x_0 & y_3-y_0 & z_3-z_0 \end{array} \right] \left[ \begin{array}{c} A \\ B \\ C \end{array} \right]$$ See above: $$= T - T_0 = \left[ \begin{array}{ccc} x-x_0 & y-y_0 & z-z_0 \end{array} \right] \left[ \begin{array}{c} A \\ B \\ C \end{array} \right]$$ Hence: $$\begin{array}{ll} x - x_0 = \xi .(x_1 - x_0) + \eta.(x_2 - x_0) + \zeta.(x_3 - x_0) \\ y - y_0 = \xi .(y_1 - y_0) + \eta.(y_2 - y_0) + \zeta.(y_3 - y_0) \\ z - z_0 = \xi .(z_1 - z_0) + \eta.(z_2 - z_0) + \zeta.(z_3 - z_0) \end{array}$$ But also: $$T - T_0 = \xi.(T_1 - T_0) + \eta.(T_2 - T_0) + \zeta.(T_3 - T_0)$$ Therefore the same expression holds for the function $T$ as well as for the coordinates $x,y,z$. This is called an isoparametric transformation. It is remarked without proof that the local coordinates $\xi,\eta,\zeta$ within a tetrahedron can be interpreted as sub-volumes, spanned by the vectors $\vec{r}_k-\vec{r}_0$ and $\vec{r}-\vec{r}_0$ where $\vec{r}=(x,y,z)$ and $k=1,2,3$. The general tetrahedron is thus mapped upon a special tetrahedron. This special tetrahedron is commonly called the parent tetrahedron. It rests in $(\xi,\eta,\zeta)$ space and it has (unit) vertices $(0,0,0)$ , $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ . The inside ($\to$ volume) of both tetrahedrons is defined by: $$\xi > 0 \quad ; \quad \eta > 0 \quad ; \quad \zeta > 0 \quad ; \quad \xi + \eta + \zeta < 1$$ Now for the volume of the general tetrahedron. It follows from the above that: $$\begin{array}{ll} dx = (x_1 - x_0)\,d\xi + (x_2 - x_0)\,d\eta + (x_3 - x_0)\,d\zeta \\ dy = (y_1 - y_0)\,d\xi + (y_2 - y_0)\,d\eta + (y_3 - y_0)\,d\zeta \\ dz = (z_1 - z_0)\,d\xi + (z_2 - z_0)\,d\eta + (z_3 - z_0)\,d\zeta \end{array}$$ Consequently: $$dx\,dy\,dz = \begin{vmatrix} (x_1 - x_0) & (x_2 - x_0) & (x_3 - x_0) \\ (y_1 - y_0) & (y_2 - y_0) & (y_3 - y_0) \\ (z_1 - z_0) & (z_2 - z_0) & (z_3 - z_0) \end{vmatrix} d\xi\,d\eta\,d\zeta$$ Upon integration, we only have to calculate for the special case: the parent tetrahedron. Christian Blatter has shown how to do such a thing: $$\iiint d\xi\,d\eta\,d\zeta = \frac{1}{6}$$ The general case follows from this: $$\iiint dx\,dy\,dz = \frac{1}{6} \begin{vmatrix} (x_1 - x_0) & (x_2 - x_0) & (x_3 - x_0) \\ (y_1 - y_0) & (y_2 - y_0) & (y_3 - y_0) \\ (z_1 - z_0) & (z_2 - z_0) & (z_3 - z_0) \end{vmatrix}$$ Let's work it out for the example at hand, where the vertices $(x_k,y_k,z_k)$ for $\;k=0,1,2,3$ , upon (renewed) calculation, are given by: $(0,2,0)$, $(8,-2,0)$ , $(0,6,0)$ , $(0,2,4)$. So the volume is: $$\frac{1}{6} \begin{vmatrix} (x_1 - x_0) & (x_2 - x_0) & (x_3 - x_0) \\ (y_1 - y_0) & (y_2 - y_0) & (y_3 - y_0) \\ (z_1 - z_0) & (z_2 - z_0) & (z_3 - z_0) \end{vmatrix} = \frac{1}{6} \begin{vmatrix} 8 & 0 & 0 \\ -4 & 4 & 0 \\ 0 & 0 & 4 \end{vmatrix} = \frac{64}{3}$$ Further Notes. The theory of the linear tetrahedron is the 3-D generalization of an analogous theory for the 2-D linear triangle:

As far as the latter reference is concerned, there exists a "closed" triangle equation in 2-D.
And there exists a "closed" tetraedron equation in 3-D as well: $$T(x,y,z) = \min( \xi , \eta , \zeta, 1 - \xi - \eta - \zeta )$$ Our "inside/outside" function $T$ is zero at the boundaries of the tetrahedron, positive inside and negative outside. Quite the same is the case with more familiar closed 3-D equations, like the one of a sphere: $\;S(x,y,z) = R^2 - (x-a)^2 - (y-b)^2 - (z-c)^2$ .
The maximum of the function $T$ is reached for $\xi = \eta = \zeta = 1 - \xi - \eta - \zeta = 1/4$ , hence at the midpoint (barycenter) of the tetrahedron.

• Oh wow, I will definitely have a detailed look on this solution when I'm back home. I hope you put that much effort into it because you enjoy yourself thinking about problems in different ways because I would never ask for so much work.. – Tesla Jun 5 '16 at 18:09
• @Sigma: Don't worry. Most of the material has been there already in LaTeX (PDF) format. Only the last part concerning the volumes is really "new". And of course I'm doing mathematics for fun, a great deal :-) – Han de Bruijn Jun 5 '16 at 19:57