Evaluating Triple integral

  • Thread starter MattL
  • Start date
  • #1
11
0
I'm having trouble with evaluating

[Triple Integral] |xyz| dx dy dz

over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1

Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?

Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,017
566
Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.

As for the parametrization,i'm sure u'll find the normal one

[tex] x=a\cos\varphi\sin\vartheta [/tex]

[tex] y=b\sin\varphi\sin\vartheta [/tex]

[tex] z=c\cos\vartheta [/tex]

pretty useful.

Daniel.
 
  • #3
saltydog
Science Advisor
Homework Helper
1,582
3
This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.
 
  • #4
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
saltydog said:
This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.
I think you missed the absolute value sign on the integrand..:wink:
 
  • #5
saltydog
Science Advisor
Homework Helper
1,582
3
arildno said:
I think you missed the absolute value sign on the integrand..:wink:
Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:

[tex]|xyz|=xyz[/tex]

That's a positive one.

However, in the octant with x<0, y>0 and z>0 we have:

[tex]|xyz|=-xyz[/tex]

And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).
 
  • #6
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
[tex]x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi[/tex]
[tex]0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi[/tex]
Thus, we may find:
[tex]dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta[/tex]
[tex]|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|[/tex]
Doing the r-integrations yield the double integral:
[tex]I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta[/tex]
We have symmetry about [tex]\phi=\frac{\pi}{2}[/tex]; thus we gain:
[tex]I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta[/tex]
We have four equal parts here, and using the part [tex]0\leq\theta\leq\frac{\pi}{2}[/tex] yields:
[tex]I=\frac{(abc)^{2}}{12}[/tex]
 
  • #7
saltydog
Science Advisor
Homework Helper
1,582
3
Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.
 
  • #8
11
0
saltydog said:
Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.
No problem.

Think I should be able to give this question a fair go now.
 
  • #9
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.

R.G. Vickson
 

Related Threads on Evaluating Triple integral

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
999
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
Top