Finding the Centroid of a Solid

  • Thread starter Thread starter dancingmonkey
  • Start date Start date
  • Tags Tags
    Centroid Solid
dancingmonkey
Messages
11
Reaction score
0

Homework Statement


Find the volume and the centroid of the solid E that lies above the cone z=√x^2+y^2 and below the sphere x^2+y^2+z^2=49.

Homework Equations




The Attempt at a Solution



My bounds were:
\theta=0 to 2\pi
\varphi=0 to \pi/4
\rho=0 to 7cos(\varphi)

So my integral was:
∫∫∫p^2sin(\varphi) d(rho) d\varphi) d(θ)

I just need help in if my bounds and integral are correct. If someone can help me with that, that would be great!
 
Physics news on Phys.org
dancingmonkey said:

Homework Statement


Find the volume and the centroid of the solid E that lies above the cone z=√x^2+y^2 and below the sphere x^2+y^2+z^2=49.

Homework Equations




The Attempt at a Solution



My bounds were:
\theta=0 to 2\pi
\varphi=0 to \pi/4
\rho=0 to 7cos(\varphi)

That last limit is wrong. What is the spherical coordinate equation of the sphere \rho=?

So my integral was:
∫∫∫p^2sin(\varphi) d(rho) d\varphi) d(θ)

I just need help in if my bounds and integral are correct. If someone can help me with that, that would be great!

And of course, once you correct the limits, that formula just gives you the volume. I think you undertand there is more to do, right?
 
So is ρ from 0 to cos(φ)? I'm still confused about this.

And yes I understand that I still have to solve the integrals :)
 
What is the equation of the sphere in spherical coordinates? Your integral must go from ρ = 0 to ρ on the sphere. So you need the equation of the sphere in spherical coordinates. Hint: It is very simple!
 
So the equation of the sphere in spherical coordinates is ρ^2=x^2+y^2+z^2. So the integral should go from p=0 to 7, right?
 
dancingmonkey said:
So the equation of the sphere in spherical coordinates is ρ^2=x^2+y^2+z^2. So the integral should go from p=0 to 7, right?

Yes! But actually the equation of the sphere in spherical coordinates is ρ = 7, not ρ = x2+y2+z2.
 
Ah! Ok I got it now, thank you so much for helping!
 
Back
Top