Question

Fun triple integral

Evaluate
\[\large \iiint_D (xyz)^m(1-x-y-z)^n dxdydz\]

\(D : x\ge 0, y\ge 0, z\ge 0, x+y+z\le 1\)

\(m,n \in \mathbb{Z^+}\)

Answer : \(\dfrac{m!^3 n!}{(3m+n+3)!}\)

Answer 1

@ganeshie8 I never did master setting up triple integrals...
but I keep getting 0<x<1-y-z
and 0<y<1 since the y intercept of x+y+z=1 is (0,1,0)
and 0<z<1 since the z intercept of x+y+z=1 is (0,0,1)

Answer 2

\[\int\limits_{0}^{1} \int\limits_0^1 \int\limits_0^{1-y-z} f(x,y,z) dx dy dz \]
but this isn;t working for me

Answer 3

```
z : 0 -->1
x : 0 --> 1-y-z
```

are looking fine... bounsd for y should be `0-->1-z`

Answer 4

oh ok that gives me the same answer as you have in the end for m=3 and n=2

Answer 5

I was just seeing if I find the same answer for some little m and n

Answer 6

nice xD i didnt verify the answer for specific cases...

Answer 7

D represents the solid region below x+y+z=1 plane in first octant

Answer 8

The actual integration is looking nasty.

Answer 9

Don't tell me any hints yet.

Answer 10

Okay :) I see how it can be nasty... here is a not so useful hint : getting a recurrence relation for the integral \(I(m,n) = \int\limits_0^a x^m(a-x)^n dx\) and evaluating it might reduce the amount of work involved...

Answer 11

\[I=\frac{n}{m+1} \cdot \frac{n-1}{m+2} \cdot \frac{n-2}{m+3} \cdot \frac{n-3}{m+4} \cdots \frac{n-(k-1)}{m+k} \int\limits _0^ax^{m+k} (a-x)^{n-k} dx\]

Answer 12

Woah! thats fast! i think we need to evaluate the last piece manually and simplify the product..

Answer 13

I needed n steps instead of k

Answer 14

\[I=\frac{n}{m+1} \cdot \frac{n-1}{m+2} \cdot \frac{n-2}{m+3} \cdot \frac{n-3}{m+4} \cdots \frac{n-(k-1)}{m+k} \int\limits\limits _0^ax^{m+k} (a-x)^{n-k} dx \\ I=\frac{n! m!}{(m+n)!}\int\limits_0^a x^{m+n}(a-x)^{0} dx = \frac{n! m!}{(m+n)!} \int\limits_0^a x^{m+n} dx\]

Answer 15

\[I=\frac{n! m!}{(m+n)!} \cdot x^{m+n+1} \frac{1}{m+n+1} |_0^a =\frac{a^{m+n+1}}{(m+n+1)!}n! m!\]

Answer 16

Nice :) actually thats a beta integral, we worked it without using the beta function xD

Answer 17

I'm glad we didn't that function
I don't know it :p

Answer 18

*
\[I(m,n) =\int\limits_0^a x^m(a-x)^n = \dfrac{m!n!}{(m+n+1)!}a^{m+n+1}\]

Answer 19

\[ \int\limits_{0}^{1-y-z}(xym)^m(1-y-z=z)^n dx \\ (ym)^m \int\limits_0^{1-y-z}x^m(1-y-z-x)^n dx \\ \text{ let } a=1-y-z \\ =(ym)^m \cdot \frac{m! n!}{(m+n+1)!} (1-y-z)^{m+n+1} \]

Answer 20

now I have to do the dy and then dz thingy

Answer 21

it the dy part should be very similar

Answer 22

looks neat, is there a typo... \((xy\color{Red}{z})^m\)

Answer 23

yes that particular m was suppose to be z

Answer 24

\[ \int\limits_{0}^{1-y-z}(xyz)^m(1-y-z-x)^n dx \\ (yz)^m \int\limits_0^{1-y-z}x^m(1-y-z-x)^n dx \\ \text{ let } a=1-y-z \\ =(yz)^m \cdot \frac{m! n!}{(m+n+1)!} (1-y-z)^{m+n+1} \]
corrections made
even that equal sign thingy

Answer 25

and that other z thingy lol

Answer 26

\[\frac{z^m m! n!}{(m+n+1)!} \int\limits_0^{1-z} y^m (1-z-y)^{m+n+1} dy\]

Answer 27

I know our a here is going to be the 1-z
but the exponent thing is going to be a little bit different
I think I just need to replace our old n with m+n+1

Answer 28

\[ \frac{z^m m! n!}{(m+n+1)!} \frac{m! (m+n+1)!}{(m+(m+n+1)+1)!} (1-z)^{m+(m+n+1)+1}\]

Answer 29

and I still need to do the last integral part

Answer 30

let me know if i made mistake in saying what I said

Answer 31

\[\frac{z^m m! n!}{(m+n+1)!} \frac{m! (m+n+1)!}{(m+(m+n+1)+1)!} (1-z)^{m+(m+n+1)+1} \\~\\~\\= \dfrac{z^mm!^2 n!}{(2m+n+2)!} (1-z)^{2m+n+2}\]

looks good to me..

Answer 32

ok just checking
because I feel like I overused that brain thing

Answer 33

\[\frac{ m! n!}{(m+n+1)!} \frac{m!(m+n+1)!}{(2m+n+2)!} \int\limits\limits_0^1 z^m (1-z)^{2m+n+2} dz \\ \]

Answer 34

\[\frac{m! n!}{(m+n+1)!} \frac{m! (m+n+1)!}{(2m+n+2)!} \frac{m! (2m+n+2)!}{(m+(2m+n+2)+1)!} =\frac{m!^3n!}{(3m+n+3)!}\]
wowzers

Answer 35

I don't think I would have been able to do it without your "useless hint"

Answer 36

Sweeet :) nothing feels better than seeing a hard problem solved neatly xD

Answer 37

Nothing feels greater than feeling like you solved a problem you couldn't solve before
thanks for the problem

Answer 38

as in me like I don't think I would have attempted to evaluate the problem
and if i did try I think I would have given up

Answer 39

From your solution it seems below is true

\[\large \iiint_D x^ay^bz^c(1-x-y-z)^n dxdydz = \dfrac{a!b!c!n!}{(a+b+c+n+3)!}\]

not sure... wil try some other time :)

Answer 40

yeah I think I will have to look at that later too
because I don't think I can use my brain anymore

Answer 41

and it does kinda look it that is true

Answer 42

I will do one small test and see

Answer 43

im not sure but i think from the integral i can tell theres a separable form to it?

Answer 44

which wed get out of the binomialexpansion

Answer 45

it does work for a=2,b=1c=3,n=4
and I know this doesn't prove for all a,b,c,n in the positive integers

Answer 46

http://www.wolframalpha.com/input/?i=%28int%28int%28int+x%5E2+y%5E1+z%5E3+%281-x-y-z%29%5E4%2C+x%3D0..1-y-z%29%2Cy%3D0..1-z%29%2Cz%3D0..1%29

http://www.wolframalpha.com/input/?i=%282*1*3*2*4*3*2%29%2F%282%2B1%2B3%2B4%2B3%29%21