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Finding total mass of hemisphere that has a variable density

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Homework Statement


Show that total mass of half solid hemisphere radius a given as x^2 + y^2 + z^2 = a^2 and z>=0 and having variable density ρ=1+(r^2)z where r is the distance of any point from the origin is given by: M = (2π(a^3))(6+3(a^3)/18 where M is total mass.


Homework Equations


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The Attempt at a Solution


My first attempt was to find the volume of the hemisphere using spherical polar coordinates and multiply this volume with the density given. Obviously this was wrong. I then tried integrating again but this time i integrated 1+(x^2 + y^2 + z^2)z
I replaced the r^2 in the density equation with a^2. I still ended up with the wrong answer. Have i applied the right method the second time round and maybe just made a mistake when integrating, or am i approaching this question the wrong way? Thanks.
 

Answers and Replies

  • #2
gabbagabbahey
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I replaced the r^2 in the density equation with a^2. I still ended up with the wrong answer.
Of course you ended up with the wrong answer; not every point inside the hemisphere is a distance [itex]a[/itex] from the origin...only the points on the outer shell of the hemisphere are that far away.

Try integrating the density over the volume of the sphere using spherical polar coordinates. What is [itex]z[/itex] in spherical polars?...What are your integration limits for [itex]r[/itex], [itex]\theta[/itex] and [itex]\phi[/itex]? What is the infinitesimal volume element (usually denoted [itex]dV[/itex]) in spherical polars?...
 
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i see! i got the answer, thank you.
 
  • #4
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I think something is wrong in your solution

My first thoughts:
since m=[itex]\rho[/itex]V

[itex]\Rightarrow[/itex] dm=Vd[itex]\rho[/itex]+[itex]\rho[/itex]dV

but in spherical : dV=r2sin([itex]\theta[/itex])dr d[itex]\theta[/itex] d[itex]\phi[/itex]
and z=r cos([itex]\theta[/itex])
[itex]\Rightarrow[/itex] dz=cos([itex]\theta[/itex]) dr - r sin([itex]\theta[/itex])d[itex]\theta[/itex]

hence, d[itex]\rho[/itex]=2 r z dr + r2 dz and substitute for dz

then substitute again in dm expression and perform the integration from
r: 0[itex]\rightarrow[/itex]a
[itex]\phi[/itex]: 0[itex]\rightarrow[/itex] 2 pi
[itex]\theta[/itex]: 0 [itex]\rightarrow[/itex] pi/2

solve it and tell me about the result :smile:
 
Last edited:
  • #5
gabbagabbahey
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I think something is wrong in your solution

My first thoughts:
since m=[itex]\rho[/itex]V

[itex]\Rightarrow[/itex] dm=Vd[itex]\rho[/itex]+[itex]\rho[/itex]dV
Huh?!:confused:

[itex]m=\rho V[/itex] is only valid for regions of uniform charge density (in which case [itex]d \rho[/itex] would be zero)

More generally, [tex]m=\int_{\mathcal{V}} \rho(\vec{r})dV[/tex].
 
  • #6
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Huh?!:confused:

[itex]m=\rho V[/itex] is only valid for regions of uniform charge density (in which case [itex]d \rho[/itex] would be zero)

More generally, [tex]m=\int_{\mathcal{V}} \rho(\vec{r})dV[/tex].
you're true ....

so [itex]dm=\rho dV[/itex] only
and the term [itex]Vd\rho[/itex] is wrong ...
 

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