Centroid of a 3D Region using Triple Integral

  • Thread starter jj2443
  • Start date
  • #1
10
0

Homework Statement


Compute the centroid of the region defined by x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex] [itex]\leq[/itex] k[itex]^{2}[/itex] and x [itex]\geq[/itex] 0 with [itex]\delta[/itex](x,y,z) = 1.

Homework Equations



[itex]\overline{x}[/itex]=[itex]\frac{1}{m}[/itex][itex]\int[/itex][itex]\int[/itex][itex]\int[/itex] x [itex]\delta[/itex](x,y,z) dV

[itex]\overline{y}[/itex]=[itex]\frac{1}{m}[/itex][itex]\int[/itex][itex]\int[/itex][itex]\int[/itex] y [itex]\delta[/itex](x,y,z) dV

[itex]\overline{z}[/itex]=[itex]\frac{1}{m}[/itex][itex]\int[/itex][itex]\int[/itex][itex]\int[/itex] z [itex]\delta[/itex](x,y,z) dV

The Attempt at a Solution


I understand that I need to integrate each of the above equations to get the x,y,z coordinates of the centroid, but how do I determine the bounds of integration?

Any help would be much appreciated! Thanks!
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,742
26
I would use spherical polar co-ordinates.
[tex]
\begin{array}{rcl}
x & = & r\sin\theta\cos\varphi \\
y & = & r\sin\theta\sin\varphi \\
z & = & r\cos\theta
\end{array}
[/tex]

with [itex]dv=r^{2}\sin\theta drd\theta d\varphi[/itex]
 

Related Threads on Centroid of a 3D Region using Triple Integral

Replies
4
Views
346
Replies
2
Views
3K
  • Last Post
Replies
2
Views
15K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
9
Views
4K
Replies
9
Views
649
Replies
9
Views
667
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
Top