Triple Integral Using Cylindrical Coordinates

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Homework Help Overview

The problem involves calculating the total mass of food in a conical container using cylindrical coordinates. The density function is given as D(r) = a/(z + 1), where 'a' is a constant and 'z' represents the height above the base of the cone.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral for mass, noting the expression for dV in cylindrical coordinates. There is confusion regarding the limits of integration for 'z', with one participant questioning the appropriateness of the limits initially proposed.

Discussion Status

There is an ongoing exploration of the correct limits for 'z', with some participants suggesting that the limits should not vary with 'r'. Clarifications are being sought regarding the implications of the conical shape versus a cylindrical shape.

Contextual Notes

Participants express uncertainty about the limits of integration, particularly for 'z', and the implications of these limits on the shape of the container being modeled. There is a recognition that the limits cannot simply be from 0 to 2, as this would imply a cylindrical shape rather than a conical one.

henryc09
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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
 
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Hi henryc09! :smile:
henryc09 said:
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:
 
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.
 
Hi henryc09! :smile:

(just got up :zzz: …)
henryc09 said:
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
henryc09 said:
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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