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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.0
dw:1347649715746:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
This can't be that hard, I must be doing something stupid.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
what are the steps you are taking?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
what is magnitude of d comparing with R ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
dw:1347649869832:dw\[V=\frac43\pi(R^3r^3)\approx4(R^3r^3)\]\[d=Rr\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
are you integrating the surface area from r to R ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
why am I integrating?
 one year ago

akash809Best ResponseYou've already chosen the best response.0
looks like u are using the right formula...expand R^3r^3 ..using formula of a^3b^3..see if u get something on expanding
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I tried and it's not seeming obvious
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
ignoring the 4 out front I want to get the r^2d part...\[r=Rd\]\[R^3(Rd)^3=3R^23Rd^2+d^3\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
\[V\approx4(3R^23Rd^2+d^3)\approx4 R^2d~~~???\]there must be a better way to see this...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
oh I should have tried difference of cubes formula maybe?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
emm..\[V=\frac{4}{3}\pi(R^3r^3)=\frac{4}{3}\pi(Rr)(R^2+rR+r^2)\]thats it max let suppose d<<R so \(r = R\) so\[V=\frac{4}{3}\pi(R^3r^3)=\frac{4}{3}\pi d(R^2+R.R+R^2)=4\pi d R^2\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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!
 one year ago

phiBest ResponseYou've already chosen the best response.0
I get 4 pi d R^2 not 4 d R^2 for R ~ r
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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/801scphysicsiclassicalmechanicsfall2010/problemsolvingandestimation/MIT8_01SC_coursenotes02.pdf
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
ah yes, the next line has the approximation as 4piR^2d
 one year ago
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