Question

Need help! find the centroid of the region inside the surfaces z = x^2 + y^2 and x^2 + y^2 + z^2 = 2

Answer 1

kinda confused any help would be great

Answer 2

The centroid of the surface itself or the centroid of the solid bounded by the two equations?

Answer 3

hmm im not really sure hold on here's the entire question:
Let W be the region inside the surfaces z = x^2 + y^2 and x^2 + y^2 + z^2 = 2. Find the centroid of W

Answer 4

so the centroid of the surface im guessing?

Answer 5

Ok. First off, you need to find the Volume of the solid bounded by the equations, and to do that you have to find where they intersect.

Answer 6

so would I plug in z=x^2+y^2 into the equation x^2+y^2+z^2=2

Answer 7

You could, but that would be hard to solve. x^2+y^2+z^2=2 is the equation of a sphere centered at the origin with radius 2^(1/2). z = x^2 + y^2 Is the equation of an elliptical parabloid opening in the positive z direction. Their intersection would be at the plane z = 1.

Answer 8

okay that makes sense

Answer 9

What you want to do next is to find the limits of the triple integration that would give you the volume of the solid.

Answer 10

okay so it would probally be easier to use spherical coordinates

Answer 11

Cylindrical would be easier. The solid is bounded below by z= x^2 + y^2 = r^2 and above by z = 1. The projection of the solid onto the xy plane is the circle x^2 + y^2 = 1, which would give r a lower limit of 0 and an upper limit of 1, and theta a lower limit of 0 and an upper limit of 2pi.

Answer 12

okay I understand the bounds but when you are trying to find the centroid what would you integrate?

Answer 13

Ah wait I've made a mistake on the bounds of z. Its bounded above by the sphere z = (2-r^2)^(1/2)

Answer 14

Well, to find the centroid of a region, you need the volume, which is just the triple integral of the limits. The centroid can then be found by the equations listed here:

http://www.physicsforums.com/showthread.php?t=244944

Answer 15

so you use the same bounds and integrate xdV, ydV, and zdV to get the 3 coordinates of the centroid

Answer 16

Also note that once you find the Volume, you don't have to find the x and y coordinates of the centroid because by symmetry they are both zero. The solid is symmetrical across both the x and y axes. You can check this by doing the integrations as well.

This just leaves the z coordinate of the centroid, so your final answer should be something along the lines of (0,0,z) where z is the z coordinate of the centroid.

Answer 17

Yes, use the same bounds and integrate over zdV, but make sure to divide the final answer by the volume.

Answer 18

\\[[V=\int\limits\limits_{0}^{2\pi}\int\limits\limits_{0}^{1}\int\limits\limits_{r^2}^{\sqrt{2-r^2}} dV\]

Answer 19

im trying to set up the integral now but I think I am still confused on something
So I set up the triple integral z dzdrdtheta using the bounds you stated above but then I have to divide by the volume of the sphere z^2+y^2+x^2=2 above the line z=1

Answer 20

okay that makes more sense now that i see it

Answer 21

Also note that dV is r dz dr dtheta, because when you convert to cylindrical you have to add the extra r in.

Answer 22

oooo thank you for reminding me I always forget that r

Answer 23

Alright I think i finally have this figured out thank you sooooo much for walking me through it sorry it took me so long but I think I understand how to do it now