Evaluate the triple integral I=∭xα−1yβ−1zγ−1dxdydz over the curve (ax)p+(by)q+(cz)r=1.
Solution
Let u=(ax)p⇒x=aup1
v=(by)q⇒y=bvq1
w=(cz)r⇒z=cwr1
and
dx=paup1−1du
dy=qbvq1−1dv
dz=rcwr1−1dw
After these substitutions, the integral becomes much lighter:
I=pqraαbβcγ∭upα−1vqβ−1wrγ−1dwdvdu evaluated over the curve u+v+w=1.
I=pqraαbβcγ∫01∫01−u∫01−u−vupα−1vqβ−1wrγ−1dwdvdu
I=pqraαbβcγγr∫01∫01−uupα−1vqβ−1(1−u−v)rγdvdu
Substitute v=(1−u)t and dv=(1−u)dt.
I=pqraαbβcγγr∫01upα−1(1−u)(qβ+rγ)∫01tqβ−1(1−t)rγdt.
Here we apply two beta functions
I=pqraαbβcγγrΓ(pα+qβ+rγ+1)Γ(pα)Γ(qβ+rγ+1)Γ(qβ+rγ+1)Γ(qβ)Γ(rγ+1).
Applying the gamma function property Γ(n+1)=nΓ(n), we arrive at the result:
I=pqraαbβcγΓ(pα+qβ+rγ+1)Γ(pα)Γ(qβ)Γ(rγ).
Check out my other notes at Proof, Disproof, and Derivation
#Calculus
#TripleIntegral
#GammaFunction
#Dirichlet
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
There are no comments in this discussion.