Symmetry in Triple Integrals: Understanding the Concept and Solving Problems

[Durdum]

Homework Statement

Why is it symmetrical thus zero?

Homework Equations

The original equation is with main focus on the first part, cause therein lies the symmetry
\int\int\int xy \quad dV + \int\int\int z^2 \quad dV
and
0 \leq z\leq 1 - x - y
See also the picture!

The Attempt at a Solution

I drew a drawing in the xy-plane and it was a line with equation y = 1-x.
So that allowed me to set up the triple integral as:

\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} xy \quad dzdydx = \frac{1}{120}

 

[BvU]

I tried calculate it hundreds of times

Could you please post your steps in detail for one of those, so we can help you ?

 

[jedishrfu]

Can you show us the triple integral for the xy term and how you evaluated it?

 

[mfb]

It would have been useful to write the problem statement here.

integrals with boundries {x, 0,1}, {y,0,1-x}, {z, 0, 1-x-y} dzdydx

If you integrate it over 1/4 of the set only, it is not zero. Integrate it over the full set (not what the image shows) and you should get zero.

 

[Durdum]

\int\int\int xy \quad dV + \int\int\int z^2 \quad dV
basically what is shown in the picture.
I chose
\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} xy \quad dzdydx
from here on it is easy peasy, I dropped it in Mathmatica and it said \frac{1}{120} so I started to cry.

It would have been useful to write the problem statement here.
If you integrate it over 1/4 of the set only, it is not zero. Integrate it over the full set (not what the image shows) and you should get zero.

What exactly do you mean with 1/4?

Yes, I plotted z = 1- x - y in Wolframalpha and I saw that the shape was symmetrical, BUT the domain is for z\geq 0. So I don't understand why it is okay. If you would say that the front in the picture is 1/4 and the back is 1/4 so a total of 1/2, then I understand but again the book wants you to calculate for z above 0, zo why multiply with 4 and not 2?

I can figure that it is an odd function because f(-x) = -f(x) but the calculation does not add up.

NOTE: I used wolframalpa to be sure that I don't mess up the algebra, so I make a mistake however I don't understand where!

 

[LCKurtz]

The integral you have calculated is correct, giving $\frac 1 {120}$. But the $xy$ domain for your pyramid looks like this:

You have only done the integral of $xy$ over the shaded portion. If you do the remaining three regions you will find that you get the same $\frac 1 {120}$ for quadrant III and $-\frac 1 {120}$ in quadrants II and IV, so the total integral is $0$. It is because of the signs of $x$ and $y$ in the various quadrants.

 

[Ray Vickson]

Homework Statement

Why is it symmetrical thus zero? I tried calculate it hundreds of times but still get a number small... but a number 1/120

Homework Equations

See the picture!

The Attempt at a Solution

integrals with boundries {x, 0,1}, {y,0,1-x}, {z, 0, 1-x-y} dzdydx

Trying to calculate the XY part.

You can see it is zero without doing any calculations.

An integral $\int\int\int_R f(x,y,x) \, dV$ is really the limit of the sums $\sum_{i} f(x_i,y_i, z_i) \Delta V$ as the number of terms $i$ goes to $\infty$ and the individual box volumes $\Delta V$ go to 0. If we fix the box sides at length $\delta > 0$, then $\Delta V = \delta^3.$ Now, for any box centered at $(x_i,y_i,z_i)$ in the first orthant, there are three other boxes centered at $(-x_i, y_i, z_i), (x_i,-y_i,z_i), (-x_i -y_i,z_i).$ The contribution of those four boxes to the sum is $[x_i y_i + (-x_i) y_i + x_i (-y_i) + (-x_i) (-y_i)] \delta^3 = 0.$

 

[Durdum]

The integral you have calculated is correct, giving $\frac 1 {120}$. But the $xy$ domain for your pyramid looks like this:
View attachment 114497
You have only done the integral of $xy$ over the shaded portion. If you do the remaining three regions you will find that you get the same $\frac 1 {120}$ for quadrant III and $-\frac 1 {120}$ in quadrants II and IV, so the total integral is $0$. It is because of the signs of $x$ and $y$ in the various quadrants.

Yes! I understand, however I only drew the first line and not the other three. Why and how could I have distilled that from the equation?

z = 1- x - y, with z = 0 in xy-plane gives y = 1- x

 

[LCKurtz]

The equation you were given was $0 \le z \le 1 - |x| - |y|$, and that right side is $1-x-y$ only in the first $xy$ quadrant where both $x$ and $y$ are positive. The equation $1 - |x| - |y| = 0$ gives the diamond shape in the $xy$ plane.

 

[Durdum]

The equation you were given was $0 \le z \le 1 - |x| - |y|$, and that right side is $1-x-y$ only in the first $xy$ quadrant where both $x$ and $y$ are positive. The equation $1 - |x| - |y| = 0$ gives the diamond shape in the $xy$ plane.

So that was my mistake! I was working on my problem for hours and couldn't find out what the problem was because I thought I did it right. The problem are the absolute values, I thought that meant it belonged in the first orthant only. However when I plotted the equation in wolframalpha.com with the absolute values I saw that it is a 4 sides pyramid.

Now I am wondering what the absolute values in $0 \le z \le 1 - |x| - |y|$ mean.

You can see it is zero without doing any calculations.

An integral $\int\int\int_R f(x,y,x) \, dV$ is really the limit of the sums $\sum_{i} f(x_i,y_i, z_i) \Delta V$ as the number of terms $i$ goes to $\infty$ and the individual box volumes $\Delta V$ go to 0. If we fix the box sides at length $\delta > 0$, then $\Delta V = \delta^3.$ Now, for any box centered at $(x_i,y_i,z_i)$ in the first orthant, there are three other boxes centered at $(-x_i, y_i, z_i), (x_i,-y_i,z_i), (-x_i -y_i,z_i).$ The contribution of those four boxes to the sum is $[x_i y_i + (-x_i) y_i + x_i (-y_i) + (-x_i) (-y_i)] \delta^3 = 0.$

This is a different approach that asks a lot of insight. I had to think about it a long time, but I think I understand it.

Thank you for helping!

 

[BvU]

This is a different approach that asks a lot of insight

An almost essential key element is the picture LC showed in #6. From that such insight is almost trivial (certainly if the last line of the problem statement spells it out).