| Author |
Message |
tamiry
science forum beginner
Joined: 28 Apr 2005
Posts: 19
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 Posted: Wed Jun 14, 2006 6:28 pm Post subject:
ball surface integral
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Hi,
this is a small exercise I always forget how to perform properly.
suppose I want to use polar coordinates to calculate ball's surface.
so the infinitesimal part of the surface is "the vertical length x the
parallel length"
((integral) r*d(teta) * r*d(phi)) and the limits are 2*PI and PI
so that's 2*PI*PI*r^2 which is unfortunately NOT 4*PI*r^2
where am I failing ?
thanks. |
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David Hartley
science forum addict
Joined: 02 Jun 2005
Posts: 86
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| Posted: Wed Jun 14, 2006 8:06 pm Post subject:
Re: ball surface integral
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In message <1150309687.766196.150610@y43g2000cwc.googlegroups.com>,
tamiry <tamir.yehuda@gmail.com> writes
| Quote: | Hi,
this is a small exercise I always forget how to perform properly.
suppose I want to use polar coordinates to calculate ball's surface.
so the infinitesimal part of the surface is "the vertical length x the
parallel length"
((integral) r*d(teta) * r*d(phi)) and the limits are 2*PI and PI
so that's 2*PI*PI*r^2 which is unfortunately NOT 4*PI*r^2
where am I failing ?
thanks.
The "parallel length" of your surface element should be |
r*sin(phi)*d(phi)
--
David Hartley |
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David Hartley
science forum addict
Joined: 02 Jun 2005
Posts: 86
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| Posted: Wed Jun 14, 2006 8:48 pm Post subject:
Re: ball surface integral
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In message <3TMT8BHJxGkEFwNO@212648.invalid>, David Hartley
<me9@privacy.net> writes
| Quote: | In message <1150309687.766196.150610@y43g2000cwc.googlegroups.com>,
tamiry <tamir.yehuda@gmail.com> writes
Hi,
this is a small exercise I always forget how to perform properly.
suppose I want to use polar coordinates to calculate ball's surface.
so the infinitesimal part of the surface is "the vertical length x the
parallel length"
((integral) r*d(teta) * r*d(phi)) and the limits are 2*PI and PI
so that's 2*PI*PI*r^2 which is unfortunately NOT 4*PI*r^2
where am I failing ?
thanks.
The "parallel length" of your surface element should be
r*sin(phi)*d(phi)
Oops, that should be |
r*sin(phi)*d(theta)
--
David Hartley |
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