Triple Integral in Cylindrical Coordinates

  • Thread starter daveyman
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
… 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
88
0
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
88
0
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
tiny-tim
Science Advisor
Homework Helper
25,836
251
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
88
0
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: 163
  • #7
tiny-tim
Science Advisor
Homework Helper
25,836
251
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 realised … 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
88
0
ah … just realised … 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
tiny-tim
Science Advisor
Homework Helper
25,836
251
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
88
0
(btw, if r is zero, then θ is meaningless :wink:)

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

Related Threads on Triple Integral in Cylindrical Coordinates

Replies
8
Views
2K
Replies
5
Views
6K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
710
Replies
2
Views
697
Replies
7
Views
7K
Replies
7
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Top