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Volume of atmosphere question

Could somebody please explain the why the volume calculated in part c, uses surface area * height rather than 4/3pi(r_1^3 - r_0^3)?
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(edited 1 month ago)
markscheme:
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Reply 2
Original post by mosaurlodon
Could somebody please explain the why the volume calculated in part c, uses surface area * height rather than 4/3pi(r_1^3 - r_0^3)?
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r1 = r0 + h
then expand the volume you get ....

Its an decent approximation, and if youre working to 3 sig fig, the +1 is lost in the noise.
(edited 1 month ago)
I agree its a decent approximation, but why use an approximation when the actual formula is probably more accurate, maybe saving time is my best guess?

I also dont really understand why surface area * height is not equal to volume but only a good approximation, it works for a prism I guess because each of the faces has a constant area, but the "face" of a sphere increases by some set amount each time?
Reply 4
Original post by mosaurlodon
I agree its a decent approximation, but why use an approximation when the actual formula is probably more accurate, maybe saving time is my best guess?
I also dont really understand why surface area * height is not equal to volume but only a good approximation, it works for a prism I guess because each of the faces has a constant area, but the "face" of a sphere increases by some set amount each time?

The working is to three 3 sig fig so r1 = 6.37*10^6 and r0 = 6.37*10^6.

As h~0.1% of the radius (roughly), just concentrating on the h*r^2 and neglecting h^2*r and h^3 is more than justified.
(edited 1 month ago)
Ok thank you that clears things up a lot more
(edited 1 month ago)
Reply 6
Original post by mosaurlodon
Ok thank you that clears things up a lot more

While the question does illustrate how limited precision (sig fig) can affect calculations, especially when you difference large numbers which are close to each other, Im not sure youd lose marks if you did your original way correctly. A couple of other observations similar to the above

Using algebra
r_1^3 - r_0^3 = (r_1 - r_0)(r_1^2 + r_1*r_0 + r_0^2) ~ h*3r_0^2
when the radii differ by a small amount h (1 in this question). So mutliply by 4/3 pi to get the volume.

Using calculus, by symmetry you can differentiate the volume in the radial/height direction so
dV/dr = 4*pi*r^2
and the volume of an infinitesimal spherical shell is that area multiplied by Deltar (=h). Altenativelly using the reverse infinitesimal argument, the volume of a spherical shell is 4*pi*r^2*Deltar, so take limits and integrate the surface area of a spherical shell with respect to r to get the volume. The volume of the thin spherical shell is what you want.
(edited 1 month ago)
Ohhh I see that is so neat - so the approximation for volume of sphere directly links with how the surface area equation was derived?
So is the following diagram a good demonstration of what you said - ill try the reverse argument as well.
Reply 8
Original post by mosaurlodon
Ohhh I see that is so neat - so the approximation for volume of sphere directly links with how the surface area equation was derived?
So is the following diagram a good demonstration of what you said - ill try the reverse argument as well.

Yes, its the same argument as the volume of a cone based on infinitesimal discs though the cone/discs is probably easier to draw than the sphere shell. To make your diagram a slightly more "accurate" you could have the missing cone in the shell where the vertex is on r0 and the base is on r1 and the base represents the difference in area between the two surfaces. Then when you shrink dr or h towards zero the cone volume(s) tend to zero and the volume / surface area interpretation becomes exact.

Just google using discs to get the volume of a cone using integration - theres plenty of videos/... out there.
(edited 1 month ago)
I was overcomplicating the volume a lot but I eventually got it



though im a bit confused by adding the missing cone.
I dont know what you mean by "where the vertex is on r0 and the base is on r1"
do you mean the vertex starts at the centre of the sphere and the base goes around the "circumference" of the sphere.
Also I dont really get how a cone links with the surface area of the sphere - maybe its because im not able to visualize it properly?
(edited 1 month ago)
Original post by mosaurlodon
I was overcomplicating the volume a lot but I eventually got it

though im a bit confused by adding the missing cone.
I dont know what you mean by "where the vertex is on r0 and the base is on r1"
do you mean the vertex starts at the centre of the sphere and the base goes around the "circumference" of the sphere.
Also I dont really get how a cone links with the surface area of the sphere - maybe its because im not able to visualize it properly?

For the cone/disc argument, you slice the cone up into discs and the area of a disc is basically pi r^2 and the height is an infinitesimal dr so the infinitesimal volume is pi r^2 dr and you sum/integrate that so
Int pi r^2 dr
which is 1/3 pi r^3. Obviously this assume a simplified cone, but it works more generally https://www.statisticshowto.com/calculus-problem-solving/volume-of-cone-proof/
and the key thing is that each disc has a height which is perpendicular to the base area so you get a stacked disc approximation which is equal to the volume of the original cone in the limit.

Its the same argument for a sphere. You have cuboids of base area 4 pi r^2 and perpendicular height dr. Then as r varies wrap them around each other to do a "staircase" appproximation of a sphere. Youd have a missing "cone" on each wrapped cuboid represesnting the cubiod approximation of the spherical shell but in the limit, the stacked/wrapped sheets tend to the volume of the sphere as the total volume of the missing cones in each layer tends to zero.
(edited 1 month ago)
I watched these 2 videos for volume of cone/sphere proved using the 'disk' method using the integral rotation around x-axis formula, but havent found one yet that uses cuboids and cones and a 'staircase' sadly

but thanks to you I am now quite clear on the reasoning behind the volume/SA of a sphere and how it links together - all starting from what I thought was a weird approximation for volume.
Original post by mosaurlodon
I watched these 2 videos for volume of cone/sphere proved using the 'disk' method using the integral rotation around x-axis formula, but havent found one yet that uses cuboids and cones and a 'staircase' sadly
but thanks to you I am now quite clear on the reasoning behind the volume/SA of a sphere and how it links together - all starting from what I thought was a weird approximation for volume.

For the cone/disc, I was just referring to the fact that the discs have width dx (in first video) that are perpedicular to the base area (parallel to the x-axis - so a staircase of successive discs) and as dx-.>0, this tends to the volume of the cone. An analogous view is that if you have cuboids of base area 4pir^2 and perpendicular height dr and wrap them around each other, then the inner/base area (r0) of each cuboid would match the sphere, but the outer area is also 4*pi*r_0^2 (not 4*pi*r_1^2) would not and youd have a small "cone" missing in each layer in the same way that the cone/disc has a small triangle missing at the top/bottom/around of each disc. In the limit, these tend to zero and the volumes match. So something like https://www.youtube.com/watch?v=kyI4kwZcl4Y&ab_channel=Mathispower4u
In their thin spherical shell (yellow) they should have a small wedge cut out as the areas of the inner and outer surface should match.

The disc method for getting the volume of a sphere that you link is the more usual way to do it, but one of archimedes claims to fame (possibly) is that he got the surface area of a sphere as 4pir^2 by making the ink with the surface area of the open ended enclosing cylinder. He also related the volume of a (hemi)sphere to the difference between the cylinder and cone volume using pythagoras so blame the greeks for all this.
(edited 1 month ago)
Oh I see, so those extra triangles that appear when you take lots of rectangles when integrating in 2D is directly analogous to extra cones that appear when you take lots of cuboids when integrating in 3D - thats kind of cool actually.
So in the video you sent, I guess the reason they didnt include those 'wedges' in their shell is that it becomes meaningless when you use lots of cuboids as the limit makes all the annoying stuff go away - thanks for all you support :smile:

and bonus thank you for the history lesson, next time while working through a problem of this type ill be sure to think of him (whether thats with respect or hatred only the future will tell)

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