Center of mass, moment of inertia for x = y^2 and y = x - 2 with density d = 3x

  • Thread starter VinnyCee
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  • #1
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Here is the problem:

A region is defined as being bounded by the parabola [tex]x = y^2[/tex] and the line [tex]y = x - 2[/tex].

The density of this region is [tex]\delta = 3x[/tex].

a) Find the center of mass.

b) Find the moment of inertia about the y-axis.

c) Find the radius of gyration about the y-axis.

Here is what I have:

[tex]M = \int_{-1}^{2}\int_{y^2}^{y + 2}\;3x\;dx\;dy = \frac{108}{5}[/tex]

[tex]M_{x} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3xy\;dx\;dy = \frac{135}{8}[/tex]

[tex]M_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3x^2\;dx\;dy = \frac{1269}{28}[/tex]

[tex]\bar{x} = \frac{\frac{1269}{28}}{\frac{108}{5}} = \frac{235}{112}\;\;and\;\;\bar{y} = \frac{\frac{135}{8}}{\frac{108}{5}} = \frac{25}{32}[/tex]

[tex]I_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} x^2 \left(3x\right)\;dx\;dy = 110.7[/tex]

[tex]R_{y} = \sqrt{\frac{110.7}{\frac{108}{5}}} \approx 2.26[/tex]

Does this look correct?
 

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Answers and Replies

  • #2
HallsofIvy
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I get what you do for M but 153/8 rather than 135/8 forMx. I get completely different answer for the others- but I may have tried to do them too fast.
 
  • #3
dextercioby
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I dunno what Halls did,but all your calculations are perfect.I double checked them with Maple...


Daniel.
 
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