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TuringTest
 3 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?
TuringTest
 3 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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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347649715746:dw

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0This can't be that hard, I must be doing something stupid.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0what are the steps you are taking?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is magnitude of d comparing with R ?

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0are you integrating the surface area from r to R ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0well 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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0why am I integrating?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I tried and it's not seeming obvious

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

TuringTest
 3 years ago
Best 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...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0oh I should have tried difference of cubes formula maybe?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0emm..\[V=\frac{4}{3}\pi(R^3r^3)=\frac{4}{3}\pi(Rr)(R^2+rR+r^2)\]thats it max let suppose d<<R so \(r = R\) so\[V=\frac{4}{3}\pi(R^3r^3)=\frac{4}{3}\pi d(R^2+R.R+R^2)=4\pi d R^2\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0This 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!

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I get 4 pi d R^2 not 4 d R^2 for R ~ r

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0that'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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0ah yes, the next line has the approximation as 4piR^2d
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