Triple Integral Using Cylindrical Coordinates

  • Thread starter henryc09
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  • #1
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Homework Statement


A conical container with radius 1, height 2 and with its base centred on the ground
at the origin contains food. The density of the food at any given point is given by
D(r) = a/(z + 1) where a is a constant and z is the height above the base.
Using cylindrical polar coordinates, calculate the total mass of food in the container.


Homework Equations





The Attempt at a Solution


ok so mass is the integral D(r)dV, and in cylindrical coordinates dV is rdrd[tex]\theta[/tex]dz

I thought that you could probably do:
[tex]\int^1_0 \,dr[/tex][tex]\int^{2\pi}_0 \,d\theta[/tex][tex]\int^{2r-2}_0 \,dz[/tex]

[tex](ra/(z+1))[/tex]

But this makes the integral very difficult and I don't think it's right. I'm pretty sure there's something wrong with my limits on dz. Any help would be appreciated
 

Answers and Replies

  • #2
tiny-tim
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Hi henryc09! :smile:
I'm pretty sure there's something wrong with my limits on dz. Any help would be appreciated

Yes, you've integrated from z = 0 to z = 2r - 2 …

why?? :confused:

z doesn't depend on r, the limits of z are the same for all r. :wink:
 
  • #3
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OK so how do I work out the limits of z? It can't just be from 0-2 because that would make it a cylinder? Still a bit confused.
 
  • #4
tiny-tim
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Hi henryc09! :smile:

(just got up :zzz: …)
I thought that you could probably do:
[tex]\int^1_0 \,dr[/tex][tex]\int^{2\pi}_0 \,d\theta[/tex][tex]\int^{2r-2}_0 \,dz[/tex]

[tex](ra/(z+1))[/tex]

But this makes the integral very difficult and I don't think it's right. I'm pretty sure there's something wrong with my limits on dz. Any help would be appreciated
OK so how do I work out the limits of z? It can't just be from 0-2 because that would make it a cylinder? Still a bit confused.

I'm sorry … somehow I read it as a cylinder. :redface:

Yes, your limits were correct.

They should lead you to an integral of rlog(2r-1) … you can use integration by parts on that.
 

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