Determine the Centroid for the volume below
- Thread starter warsatan
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Re: Determine the Centroid
warsatan said:I took Cal a while back and now taking Static, not sure how to do this, can anyone help me out? thanks alot.
hmm, not sure what's the proper way to display an file image in this room, below is the file attached that i scanned. [\quote]
Sorry, youre picture didn't show up.
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- Feb 4, 2004
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Yes, but the image will be "scaled down to fit", because the image you've uploaded is so huge -- it's a full meg. It might help if you uploaded a smaller version.warsatan said:
For the tutors: The text of the exercise is as follows:
The volume shown in the graphic looks like an ice-cream cone, with the z-axis passing upward from the base of the cone and out through the top of the scoop of ice cream.
Eliz.
it's only 160kb. anyhow, i saved it in a word documents, and here is the link to download. If you have a chance, please take a look. Thanks
http://www.yousendit.com/transfer.php?a ... E20BADA6C3
http://www.yousendit.com/transfer.php?a ... E20BADA6C3
Image shack will only work up to 1.5MB. You can resize your pictures. I have to do it all the time.
Anyway, I think I know what your image looks like.
Try spherical coordinates, though, there are other ways.
It's been a while since I messed with these. I hope this helps.
It sounds like you have to find the volume and the centroid of the solid bounded above by the sphere \(\displaystyle x^{2}+y^{2}+z^{2}=1\) and below by the cone \(\displaystyle z=\sqrt{x^{2}+y^{2}}\)
Try:
\(\displaystyle \L\\V=\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{3}}\int_{0}^{1}{\rho}^{2}sin({\phi})\, d{\rho}\, d{\phi}\, d{\theta}\)
That's the volume. Once you know that you can find the centroid.
By symmetry, the centroid is on the z-axis, so \(\displaystyle \overline{x}\) and \(\displaystyle \overline{y}\) =0
Since \(\displaystyle z={\rho}sin({\phi})\).
and use \(\displaystyle \overline{z}=\frac{1}{V}\int\int\int\, z\, dV\)
\(\displaystyle \L\\\frac{1}{V}\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{3}}\int_{0}^{1}\, ({\rho}sin({\phi})){\rho}^{2}sin({\phi})\, d{\rho}\, d{\phi}\, d{\theta}\)
Anyway, I think I know what your image looks like.
Try spherical coordinates, though, there are other ways.
It's been a while since I messed with these. I hope this helps.
It sounds like you have to find the volume and the centroid of the solid bounded above by the sphere \(\displaystyle x^{2}+y^{2}+z^{2}=1\) and below by the cone \(\displaystyle z=\sqrt{x^{2}+y^{2}}\)
Try:
\(\displaystyle \L\\V=\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{3}}\int_{0}^{1}{\rho}^{2}sin({\phi})\, d{\rho}\, d{\phi}\, d{\theta}\)
That's the volume. Once you know that you can find the centroid.
By symmetry, the centroid is on the z-axis, so \(\displaystyle \overline{x}\) and \(\displaystyle \overline{y}\) =0
Since \(\displaystyle z={\rho}sin({\phi})\).
and use \(\displaystyle \overline{z}=\frac{1}{V}\int\int\int\, z\, dV\)
\(\displaystyle \L\\\frac{1}{V}\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{3}}\int_{0}^{1}\, ({\rho}sin({\phi})){\rho}^{2}sin({\phi})\, d{\rho}\, d{\phi}\, d{\theta}\)