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

- chestercat

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- TuringTest

|dw:1347649715746:dw|

- TuringTest

This can't be that hard, I must be doing something stupid.

- amistre64

what are the steps you are taking?

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## More answers

- anonymous

what is magnitude of d comparing with R ?

- TuringTest

|dw:1347649869832:dw|\[V=\frac43\pi(R^3-r^3)\approx4(R^3-r^3)\]\[d=R-r\]

- amistre64

are you integrating the surface area from r to R ?

- TuringTest

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

- TuringTest

why am I integrating?

- anonymous

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

- TuringTest

I tried and it's not seeming obvious

- TuringTest

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

- TuringTest

R^2d I mean...

- TuringTest

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

- TuringTest

oh I should have tried difference of cubes formula maybe?

- anonymous

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<

- TuringTest

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!

- anonymous

np

- phi

I get 4 pi d R^2 not 4 d R^2
for R ~ r

- TuringTest

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

- phi

oh... see mukushia

- TuringTest

ah yes, the next line has the approximation as 4piR^2d

- phi

yep, a typo

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