- #1

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He says:

[tex]dV=πy^{2}dz[/tex]

However, if we're finding the surface area of the sphere:

[tex]dA=2πRdz≠2πydz[/tex]

If we cannot use [tex]dA=2πydz[/tex], how come [tex]dV=πy^{2}dz[/tex] is still applicable?

- Thread starter greswd
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- #1

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He says:

[tex]dV=πy^{2}dz[/tex]

However, if we're finding the surface area of the sphere:

[tex]dA=2πRdz≠2πydz[/tex]

If we cannot use [tex]dA=2πydz[/tex], how come [tex]dV=πy^{2}dz[/tex] is still applicable?

- #2

Doc Al

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That's the volume of those circular disks.He says:

[tex]dV=πy^{2}dz[/tex]

What does finding the surface area of the sphere have to do with this?

- #3

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What does finding the surface area of the sphere have to do with this?

If I want to find the MoI of a hollow sphere.

Based on the same assumptions [tex]dA=2πydz[/tex], it should be like that, but it's not .

I'm trying to calculate the MoIs of some hollow objects, not sure what kind of area formula I should use.

- #4

Doc Al

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Not sure what you mean by "the same assumptions". To find the surface area of that ring, you need the arc length, not the vertical distance (dz).If I want to find the MoI of a hollow sphere.

Based on the same assumptions [tex]dA=2πydz[/tex], it should be like that, but it's not .

- #5

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Yup, that's the perplexing part of the formula [tex]A=2πRz[/tex]Not sure what you mean by "the same assumptions". To find the surface area of that ring, you need the arc length, not the vertical distance (dz).

As z tends to zero, that particular segment of a sphere should tend toward a segment of a cone. And for a cone, when z tends to zero

[tex]A→2πrz[/tex]

The difference between the conical area and the area of a cylinder wrapped around it should decrease until they are equal when z tends to zero.

- #6

Doc Al

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Again, it's the length of that side that must be used. And that length is not simply z. (It's at an angle.)Yup, that's the perplexing part of the formula [tex]A=2πRz[/tex]

As z tends to zero, that particular segment of a sphere should tend toward a segment of a cone. And for a cone, when z tends to zero

[tex]A→2πrz[/tex]

Their ratio remains finite. They aren't the same.The difference between the conical area and the area of a cylinder wrapped around it should decrease until they are equal when z tends to zero.

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Is this for the moment of inertia of a sphere?

- #8

Doc Al

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A spherical shell.Is this for the moment of inertia of a sphere?

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- #10

Doc Al

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Yes, that would be correct if he were finding the moment of inertia of a solid sphere, but he's not.

- #11

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That sounds correct.Again, it's the length of that side that must be used. And that length is not simply z. (It's at an angle.)

But why does this not apply to volume integrals?

- #12

Doc Al

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Any slant with the side will be a second order correction to the radius and will have no effect on the volume element.But why does this not apply to volume integrals?

The volume is the area times thickness:

dV = area * dz = πr

If we vary the radius by dr, we get:

dV = π(r + dr)

dV = πr

The higher order differentials can be ignored.

- #13

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So ignoring the higher order differentials, we still get an exact answer?Any slant with the side will be a second order correction to the radius and will have no effect on the volume element.

The volume is the area times thickness:

dV = area * dz = πr^{2}dz

If we vary the radius by dr, we get:

dV = π(r + dr)^{2}dz = π(r^{2}+ 2rdr + dr^{2})dz

dV = πr^{2}dz + [STRIKE]2πrdrdz[/STRIKE] + [STRIKE]πdr^{2}dz[/STRIKE]

The higher order differentials can be ignored.

What if the external area is perimeter times thickness:

dA = perimeter * dz =2πrdz

Varying the radius by dr, we get:

dA = 2π(r+dr)dz =(2πr+2πdr)dz = 2πrdz + [STRIKE]2πdrdz

[/STRIKE]

- #14

Doc Al

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Yes, that's how it's done. In the limit as they go to zero, (dx)So ignoring the higher order differentials, we still get an exact answer?

But the surface area isWhat if the external area is perimeter timesthickness:

- #15

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ah, that makes sense.Yes, that's how it's done. In the limit as they go to zero, (dx)^{2}can be ignored compared to dx.

But the surface area isnotperimeter times thickness; it's perimeter times the length of the side.

but then volume can also be considered area times some funny ratio and not just the vertical height dz

- #16

Doc Al

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You can express the length of the side in terms of the height. But that will also depend on the angle.but then volume can also be considered area times some funny ratio and not just the vertical height dz

- #17

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yeah it will. so you've already explained why we can use dz for volume integrals, by ignoring the higher order differentials.You can express the length of the side in terms of the height. But that will also depend on the angle.

but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?

I'm afriad my mathematical understanding is really sketchy.

- #18

AlephZero

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No. Make a drawing of a cut through the center of the sphere. The cut surface is a circle radius r. You have a little right-angled triangle, with the horizontal side length dr, the vertical side length dz, and the slant (the actual surface) is length ##\sqrt{dr^2 + dz^2}## by Pythagoras's theorem.but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?

Of course you can use trig and get the length in terms of an angle, if you want.

- #19

Doc Al

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No. See AlephZero's response. (He beat me to it. )but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?

- #20

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I get it, but, I don't understand why aleph's method doesn't apply to volume integrals.No. See AlephZero's response. (He beat me to it. )

- #21

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Let's say I'm trying to find the circumference and area of a circle by using rectangles to approximate.

As the no. of rectangles approaches infinity, their combined height will always be equal to the diameter, and will never approximate the circumference.

However, the combined area of those rectangles will approximate the area of the circle.

This is why dz works for volume and not area integrals.

It's a geometric proof, but how do we prove it mathematically?

- #22

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Revisiting an old thread, does the proof lie in line integrals?

- #23

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Yes. Integration rocks !! As always. Also, as I can see by your thread, you are inquisitive of finding moment of inertia of a solid sphere, right ? I did not follow the hyperphysics derivation, but I derive it using moment of inertia of hollow sphere, which is 2MRRevisiting an old thread, does the proof lie in line integrals?

I am sorry. You ought to find moment of inertia of hollow sphere ?

There are two ways:

1. Integration (difficult)

2. Coordinate geometry (easy!).. which I think you have been told here.

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