Triple Integral in Cylindrical Coordinates

In summary, we are trying to find the mass and center of mass of a solid bounded by a paraboloid and a plane. We use cylindrical coordinates to find the mass, which is equal to 1/8 of the constant density K multiplied by the square of the upper limit of the integration for z. However, when trying to find the r component of the center of mass, we realize that we cannot do so in cylindrical coordinates and must convert to x and y coordinates. The z coordinate for the center of mass can still be found in cylindrical coordinates. In the end, it is determined that the r and theta components of the center of mass are both equal to zero.
  • #1
daveyman
88
0

Homework Statement


Find the mass and center of mass of the solid S bounded by the paraboloid [tex]z=4x^2+4y^2[/tex] and the plane [tex]z=a\;\;(a>0)[/tex] if S has constant density K.


Homework Equations


In cylindrical coordinates, [tex]x^2+y^2=r^2[/tex].


The Attempt at a Solution


In order to find the mass, I tried
[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta[/tex]
This didn't seem to work, though.
 
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  • #2
daveyman said:
… I tried
[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta[/tex]
This didn't seem to work, though.

Hi daveyman! :smile:

Hint: the volume element isn't drdzdθ. :wink:
 
  • #3
Okay, so this is the mass:

[tex]\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}\text{Kr}\;\;drdzd\theta = \frac{1}{8} a^2 K \pi[/tex]
 
Last edited:
  • #4
Now that I have the mass (m), I want to find the r component of the center of mass. I tried

[tex]\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta [/tex]

This is incorrect. What am I doing wrong?
 
Last edited:
  • #5
daveyman said:
Okay, so now that I have the mass (m), I want to find the r component of the center of mass. I tried

[tex]\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta [/tex]

This is incorrect. What am I doing wrong?

That should come out as 0, which is the r component of the centre of mass. :wink:
 
  • #6
Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
I've included the output as an attachment.
 

Attachments

  • Output.pdf
    81.3 KB · Views: 246
  • #7
daveyman said:
Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
I've included the output as an attachment.

ah … just realized … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

(but the z coordinate will be ok :smile:)
 
  • #8
tiny-tim said:
ah … just realized … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

(but the z coordinate will be ok :smile:)

Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?
 
  • #9
daveyman said:
Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?

I think you're perfectly entitled to say that it's obvious. :smile:

(btw, if r is zero, then θ is meaningless :wink:)
 
  • #10
tiny-tim said:
(btw, if r is zero, then θ is meaningless :wink:)

Very true. I think that's it. Thank you!
 

Related to Triple Integral in Cylindrical Coordinates

1. What is a triple integral in cylindrical coordinates?

A triple integral in cylindrical coordinates is a mathematical tool used to calculate the volume of a three-dimensional object with cylindrical symmetry. It involves integrating a function over a region in three-dimensional space, taking into account the cylindrical coordinates (radius, angle, and height) of the object.

2. How is a triple integral in cylindrical coordinates different from a regular triple integral?

A triple integral in cylindrical coordinates is different from a regular triple integral in that it takes into account the cylindrical coordinates of the object being integrated. This means that the bounds of integration are expressed in terms of radius, angle, and height, rather than just x, y, and z coordinates.

3. What are the advantages of using cylindrical coordinates in a triple integral?

The main advantage of using cylindrical coordinates in a triple integral is that it simplifies the integration process for objects with cylindrical symmetry. This is because the bounds of integration are typically easier to determine and the integrand can often be expressed in terms of simpler functions.

4. What is the formula for a triple integral in cylindrical coordinates?

The formula for a triple integral in cylindrical coordinates is ∭f(r,θ,z)dV = ∫∫∫f(r,θ,z)rdrdθdz, where r is the radius, θ is the angle, and z is the height, and dV is the infinitesimal volume element.

5. How can I convert a triple integral in cylindrical coordinates to Cartesian coordinates?

To convert a triple integral in cylindrical coordinates to Cartesian coordinates, you can use the following formulas: x = rcosθ, y = rsinθ, and z = z. Then, you can change the bounds of integration accordingly. It is important to note that the Jacobian determinant |J| = r must be included in the integral when converting from cylindrical to Cartesian coordinates.

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