Triple Integral Using Cylindrical Coordinates

In summary, the container contains a certain amount of food at every point, the mass of the food is given by the equation D(r) = a/(z+1), and the limits of z are the same for all r.
  • #1
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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  • #2
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:
 
  • #3
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
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.
 

FAQ: Triple Integral Using Cylindrical Coordinates

1. What is a triple integral using cylindrical coordinates?

A triple integral using cylindrical coordinates is a mathematical calculation used to find the volume within a three-dimensional region bounded by a cylindrical surface and two perpendicular planes. It involves integrating a function over a cylindrical volume using the variables of radius, height, and angle.

2. When is it best to use cylindrical coordinates for a triple integral?

Cylindrical coordinates are best used for a triple integral when the region being integrated over has a cylindrical symmetry, meaning it is circular or cylindrical in shape. This allows for simplification of the integral and makes the calculation more efficient.

3. How do you convert a triple integral from rectangular to cylindrical coordinates?

To convert a triple integral from rectangular coordinates to cylindrical coordinates, you must substitute the variables of x, y, and z with the corresponding cylindrical variables of radius (ρ), angle (θ), and height (z). You must also adjust the limits of integration to match the new coordinate system.

4. What are the advantages of using cylindrical coordinates for a triple integral?

Using cylindrical coordinates for a triple integral can have several advantages, including simplifying the integral and making the calculation more efficient, as well as allowing for easier visualization and understanding of the region being integrated over.

5. Can a triple integral using cylindrical coordinates be used to solve real-world problems?

Yes, a triple integral using cylindrical coordinates can be used to solve real-world problems in physics, engineering, and other fields. It can be used to calculate volumes of objects with cylindrical symmetry, such as pipes or tanks, and to solve problems involving flow and rotation, among others.

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