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Find the centroid of the solid bounded below by the cone

  • Thread starter wilcofan3
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Homework Statement



Find the centroid of the solid bounded below by the cone [tex]z = \sqrt{3(x^2+y^2)}[/tex] and bounded above the sphere [tex]x^2+y^2+z^2=36[/tex].

Homework Equations



Let G be the given solid and denote its volume by [tex]V_{G}=\int \int \int_{G} 1 dV.[/tex]

[tex]\frac{\bar{x}= \int \int \int_{G} x dV}{V_{G}}[/tex],[tex]\frac{\bar{y}= \int \int \int_{G} y dV}{V_{G}}[/tex],[tex]\frac{\bar{z}= \int \int \int_{G} z dV}{V_{G}}[/tex]

The Attempt at a Solution



I know I need to use spherical coordinates making the solid G given by:

[tex]x = r sin \phi cos \theta[/tex]
[tex]y = r sin \phi sin \theta[/tex]
[tex]z = r cos \phi[/tex]

Now, for the limits of integration (the area where I struggle most with these integrals!), I think [tex]0 \leq r \leq 6[/tex] and [tex]0 \leq \theta \leq 2\pi[/tex] [tex]0 \leq \phi \leq \frac{3\pi}{4}[/tex]

This is stupid, but where do I go from here if this is all correct?
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi wilcofan3! :smile:

(try using the X2 tag just above the Reply box :wink:)
Find the centroid of the solid bounded below by the cone [tex]z = \sqrt{3(x^2+y^2)}[/tex] and bounded above the sphere [tex]x^2+y^2+z^2=36[/tex].

I know I need to use spherical coordinates making the solid G given by:

[tex]x = r sin \phi cos \theta[/tex]
[tex]y = r sin \phi sin \theta[/tex]
[tex]z = r cos \phi[/tex]

Now, for the limits of integration (the area where I struggle most with these integrals!), I think [tex]0 \leq r \leq 6[/tex] and [tex]0 \leq \theta \leq 2\pi[/tex] [tex]0 \leq \phi \leq \frac{3\pi}{4}[/tex]

This is stupid, but where do I go from here if this is all correct?
(erm :redface: … 3π/4 is 135º)

Now multiply by the volume element (ie the r θ φ equivalent of dxdydz), and integrate. :wink:
 

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