Finding the Mass of a Solid Using Triple Integrals

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Homework Help Overview

The problem involves finding the mass of a solid defined by the region outside the sphere described by the equation x² + y² + z² = 1 and inside the surface given by x² + y² + z² = 2z. The subject area pertains to triple integrals and spherical coordinates.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to determine the boundaries for the radius in spherical coordinates but expresses confusion regarding the second surface's shape and its implications for the integration limits.
  • Some participants question the assumption that the second surface is a sphere and suggest exploring its actual geometric representation.
  • There is a discussion about the nature of the second surface, with attempts to visualize it through level curves and comparisons to known shapes.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the geometric interpretation of the surfaces involved. Some guidance has been offered regarding the setup of the integral, but no consensus has been reached on the exact boundaries for integration.

Contextual Notes

There is a noted confusion about the nature of the second surface, which is initially misidentified as a sphere. The original poster acknowledges potential oversights in their understanding of the problem setup.

FunkyDwarf
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Hey guys!

Just a quick question. As usual I am sure I am missing something stupid but if you can help me get my head around it id appreciate it (ie sorry if it seems a mundane error)

Homework Statement


Find the mass of the solid T outside the sphere [tex]x^2 + y^2 + z^2 = 1[/tex]
and inside the sphere [tex]x^2 +y^2 + z^2 = 2z[/tex]

Homework Equations


Ok clearly given theyre spheres spherical coordinates is the go, but i kinda hit a stang finding the boundary for the radius or rho.

The Attempt at a Solution


Clearly the first sphere has radius one, but what about the 2nd one? Given its related to 2z it would seem more of an ellipse type thing...

given that the LHS is r^2 i used the relation z = rcos(phi) which gives the upper bound of r as cos(phi)

Does that sound right?

Cheers
-G
 
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[tex]x^2+y^2+z^2=2z[/tex] isn't a sphere, so don't try and make it one.

you might want to start by noting if z<1/2, the x-y cross section of the outer shell is smaller than the x-y cross section of the inner shell... this gives you one bound for z. Try and find the other one.
 
Ah ok fair nuff, will do. Yeh i thought it wasnt a sphere, but the question stated it as one so guess its a misprint. thanks, will give it a go.

EDIT: Ok tried drawing some level curves...not sure theyre right. am i correct in thinking the 2nd function (2z) looks something like a y=x^2 curve but rotated around the y-axis (centered about z here though)?

if so how can we use that as our upper bound because it is unbounded (in the z variable anyway) ...

sorry if seems like I am making stupid oversights :S

cheers
-G
 
Last edited:
Office_Shredder said:
[tex]x^2+y^2+z^2=2z[/tex] isn't a sphere, so don't try and make it one.

[tex]x^2+y^2+z^2=2z[/tex]
[tex]x^2+y^2+z^2-2z=0[/tex]
[tex]x^2+y^2+z^2-2z+1=1[/tex]
[tex]x^2+y^2+\left(z-1\right)^2=1[/tex]

Looks a whole lot like a sphere of radius 1, centered at [itex]<0,0,1>[/itex] to me.

This looks very much like a question you're supposed to answer by setting up an integral.
 
Knew it was something simple :) guess i should have done a few more iterations of th level surface huh

thanks mate
-G
 

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