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TuringTest

  • 2 years ago

Why can I not seem to show that the volume of a spherical shell is\[V\approx4R^2d\]where R is the outer radius and d is the thickness?

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  1. TuringTest
    • 2 years ago
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    |dw:1347649715746:dw|

  2. TuringTest
    • 2 years ago
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    This can't be that hard, I must be doing something stupid.

  3. amistre64
    • 2 years ago
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    what are the steps you are taking?

  4. mukushla
    • 2 years ago
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    what is magnitude of d comparing with R ?

  5. TuringTest
    • 2 years ago
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    |dw:1347649869832:dw|\[V=\frac43\pi(R^3-r^3)\approx4(R^3-r^3)\]\[d=R-r\]

  6. amistre64
    • 2 years ago
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    are you integrating the surface area from r to R ?

  7. TuringTest
    • 2 years ago
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    well it's not supposed to be negligible. this is supposed to be a rough model of the earth were it covered completely in water, but I don't think I am to assume that d is that small that we can drop it from the calculations

  8. TuringTest
    • 2 years ago
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    why am I integrating?

  9. akash809
    • 2 years ago
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    looks like u are using the right formula...expand R^3-r^3 ..using formula of a^3-b^3..see if u get something on expanding

  10. TuringTest
    • 2 years ago
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    I tried and it's not seeming obvious

  11. TuringTest
    • 2 years ago
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    ignoring the 4 out front I want to get the r^2d part...\[r=R-d\]\[R^3-(R-d)^3=3R^2-3Rd^2+d^3\]

  12. TuringTest
    • 2 years ago
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    R^2d I mean...

  13. TuringTest
    • 2 years ago
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    \[V\approx4(3R^2-3Rd^2+d^3)\approx4 R^2d~~~???\]there must be a better way to see this...

  14. TuringTest
    • 2 years ago
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    oh I should have tried difference of cubes formula maybe?

  15. mukushla
    • 2 years ago
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    emm..\[V=\frac{4}{3}\pi(R^3-r^3)=\frac{4}{3}\pi(R-r)(R^2+rR+r^2)\]thats it max let suppose d<<R so \(r = R\) so\[V=\frac{4}{3}\pi(R^3-r^3)=\frac{4}{3}\pi d(R^2+R.R+R^2)=4\pi d R^2\]

  16. TuringTest
    • 2 years ago
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    This is why I'm practicing this kind of reasoning, I would not have thought to use \(r=R\) at that point. The MIT problems are tough; estimating the mass of water on the earth... Thanks mukushla!

  17. mukushla
    • 2 years ago
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    np

  18. phi
    • 2 years ago
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    I get 4 pi d R^2 not 4 d R^2 for R ~ r

  19. TuringTest
    • 2 years ago
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    that's what I was about to say... maybe there is a typo on the doc; there is a weird space after the 4 (check 2.3.5) http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/problem-solving-and-estimation/MIT8_01SC_coursenotes02.pdf

  20. phi
    • 2 years ago
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    oh... see mukushia

  21. TuringTest
    • 2 years ago
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    ah yes, the next line has the approximation as 4piR^2d

  22. phi
    • 2 years ago
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    yep, a typo

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