1
$\begingroup$

I am uncertain about what I might be overlooking, but I am attempting to determine the volume of a shape with the following boundaries: $0<z<y<x<1$ and $y+z<1$ .

since in 2D for $z<1-y$ and $z<y$ , the integral is calculated as follows:

$$ \int_{0}^{0.5} \int_{0}^{z} dydz + \int_{0.5}^{1} \int_{0}^{1-z} dydz = \frac{1}{4} $$ and for the condition 0<z<y<x<1 :
$$ \int_{0}^{1} \int_{z}^{1} \int_{y}^{1} dxdydz = \frac{1}{6} $$

so I tried to apply the same principal here and I get :

$$ \int_{0}^{0.5} \int_{z}^{1} \int_{y}^{1} dxdydz + \int_{0.5}^{1} \int_{0}^{1-z} \int_{1-y}^{1} dxdydz $$

However this is incorrect, I would appreciate any insights or thoughts on this matter.

$\endgroup$
2
  • $\begingroup$ If you have access to the correct answer some users [like myself] might like to know that in order to check an approach. $\endgroup$
    – coffeemath
    Commented Nov 19, 2023 at 20:12
  • $\begingroup$ I actually don't have it yet, but I will definitely post it down here. The original question was a joint probability problem, which has transformed into a triple integral problem based on the above description. I believe there is another approach involving the calculation of marginal distribution, which leads to a simpler integral form. $\endgroup$
    – Elias
    Commented Nov 21, 2023 at 7:17

2 Answers 2

0
$\begingroup$

The upper bound on $z$ is $\min(y,1-y)$ which has the formula $$h(y)=\frac12 - |y-\frac12|.$$ So the volume integral is $$ \int_0^1 \int_0^x \int_0^{h(y)} 1\ dz\ dy\ dx.$$ I put that in a symbolic calculator and it gave $\frac18 .$

Added note: I found that wolfram alpha can use min(y,1-y) for the upper bound of inner integral. If interested here's link to wolframs calculation. I dont have access to the step by step part.

https://www.wolframalpha.com/input?i=int_0%5E1+int_0%5Ex+int_0%5Emin%28y%2C1-y%29+1+dz+dy+dx

$\endgroup$
0
$\begingroup$

The region defined by $0 < z < y < x < 1$ will be the interior of the tetrahedron with vertices $$(0,0,0), (1,0,0), (1,1,0), (1,1,1).$$ We can see this is true by observing that $\inf x = \inf y = \inf z = 0$, and $\sup x = \sup y = \sup z = 1$; i.e., the region is contained within the unit cube. Then the inequality $x > y$ implies the region projects onto the $xy$-plane the triangle with vertices $(0,0,0)$, $(1,0,0)$, $(1,1,0)$, and similar reasoning for the other projections in the $xz$- and $yz$-planes yields the desired region.

Then add in the condition $y+z < 1$, the boundary of which is a plane that passes through the points $(0,1,0)$ and $(0,0,1)$. This "cuts" the tetrahedron so that the base in the $xy$-plane remains the same, but the projection in the $yz$-plane is now the function $$z(y) = \min(y, 1-y), \quad 0 < y < 1.$$ So now the region has the following vertices: $$O = (0,0,0), \quad A = (1,0,0), \quad B = (1,1,0), \quad C = (1,1/2,1/2), \quad D = (1/2,1/2,1/2),$$ and the faces $$\triangle OAB, \quad \triangle ABC, \quad \triangle OBD, \quad \triangle BCD, \quad OACD$$ where $OACD$ is a trapezoid. Slicing this region along a plane at $z = z_0$ for $0 < z_0 < 1/2$ yields the trapezoid with vertices $$(z_0, z_0, z_0), (1, z_0, z_0), (1, 1-z_0, z_0), (1-z_0, 1-z_0, z_0).$$

Therefore, the region may be integrated as $$\int_{z=0}^{1/2} \int_{y=z}^{1-z} \int_{x=y}^1 \, dx \, dy \, dz.$$

enter image description here

$\endgroup$
1
  • $\begingroup$ Great visualization! How did you manage to create it? I used Python Plotly for an interactive 3D plot, but with limited success in gaining more intuition about it. Secondly, if I add two more constraints like y < Y and z < Z and try to calculate this new integral parametrically, what would be your approach? $\endgroup$
    – Elias
    Commented Nov 22, 2023 at 12:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .