Finding the Centroid of a Solid

  • #1

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:
[itex]\theta[/itex]=0 to 2[itex]\pi[/itex]
[itex]\varphi[/itex]=0 to [itex]\pi[/itex]/4
[itex]\rho[/itex]=0 to 7cos([itex]\varphi[/itex])

So my integral was:
∫∫∫p^2sin([itex]\varphi[/itex]) d(rho) d[itex]\varphi[/itex]) 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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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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:
[itex]\theta[/itex]=0 to 2[itex]\pi[/itex]
[itex]\varphi[/itex]=0 to [itex]\pi[/itex]/4
[itex]\rho[/itex]=0 to 7cos([itex]\varphi[/itex])
That last limit is wrong. What is the spherical coordinate equation of the sphere [itex]\rho=?[/itex]

So my integral was:
∫∫∫p^2sin([itex]\varphi[/itex]) d(rho) d[itex]\varphi[/itex]) 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?
 
  • #3
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 :)
 
  • #4
LCKurtz
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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!
 
  • #5
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?
 
  • #6
LCKurtz
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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.
 
  • #7
Ah! Ok I got it now, thank you so much for helping!
 

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