Finding centroid of a solid

[Phil Frehz]

Homework Statement

Use cylindrical coordinates to find the centroid of the solid.

The solid that is bounded above by the sphere x² + y² + z² = 2

and below by z = x² + y²

Homework Equations

x = rcos(theta)

y= rsin(theta)[/B]

The Attempt at a Solution

I am having trouble trying to find the limits of integration for r.

I have been able to get the sphere in terms of r as z = (2-r²)^1/2 and the paraboloid as z = r²

I understand that in cylindrical coordinates the region is occupied from 0<theta<2pi and r originates from origin n to the uppermost part of the sphere.

I also found that when you equate both surface z equations. you get r² = (2-r²)^1/2 . Solving for r you get radical 2 and 1. r being the furthest distance would then have the upper limit of 1. I'm not sure if this is the correct method of solving.[/B]

 

[RUber]

First off, $\sqrt 2>1$ and $\sqrt 2$ doesn't seem to solve your equation.

Go with z=1 for the intersection point, so z is defined by the sphere above 1<z <2 and as the paraboloid for 0<z <1.

 

[RUber]

See the attached image, this is the shape you are dealing with (revolved around the z axis to form a volume). Clearly, your centroid will be at x=0, y=0. How will you find the centroid in the z direction?

 

[Phil Frehz]

See the attached image, this is the shape you are dealing with (revolved around the z axis to form a volume). Clearly, your centroid will be at x=0, y=0. How will you find the centroid in the z direction?
View attachment 86271

So what I did was found out the value of z at the intersection point by equating both surfaces in terms of r, which led z to be either -2 or 1. When inputting z =1 i get r to also be 1

The centroid in the z direction will thus be ₀ ∫^2pi ₀ ∫¹ _r² ∫ ^√(2-r2) z ⋅ r dz dr dθ all divided by the volume of the region

 

[RUber]

So what I did was found out the value of z at the intersection point by equating both surfaces in terms of r, which led z to be either -2 or 1. When inputting z =1 i get r to also be 1

Do you mean you solved $r^2 = (2-r^2)^{1/2}$ as a quadratic to find $r^2 = 1 \text{ or } -2$ and then discarded the imaginary solution?

The centroid in the z direction will thus be ₀ ∫^2pi ₀ ∫¹ _r² ∫ ^√(2-r2) z ⋅ r dz dr dθ all divided by the volume of the region

Any troubles with these integrals?

 

[Phil Frehz]

Do you mean you solved $r^2 = (2-r^2)^{1/2}$ as a quadratic to find $r^2 = 1 \text{ or } -2$ and then discarded the imaginary solution?


Any troubles with these integrals?

Yes I discarded the imaginary solution, I have no problem solving the integrals. I only had trouble identifying the upper limit of r. Would you say that the way I solved for r was a correct solution?

 

[RUber]

That is the correct way to do it.
What you did wrong on your original work posted in the image was to go from
$r^4 + r^2 =2$ to r =1 or r = sqrt(2); you should have gotten r =1 or r = sqrt(-2).

 

[Phil Frehz]

That is the correct way to do it.
What you did wrong on your original work posted in the image was to go from
$r^4 + r^2 =2$ to r =1 or r = sqrt(2); you should have gotten r =1 or r = sqrt(-2).

Ah I overlooked the proper solution for r. Thanks for taking the time to help me out!