Here's a derviation from HyperPhysics: 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?
That's the volume of those circular disks. 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.
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).
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] 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.
Again, it's the length of that side that must be used. And that length is not simply z. (It's at an angle.) Their ratio remains finite. They aren't the same.
I'm pretty sure it's for a full sphere, that dI there represents the moment of inertia of a single disk at height z, and then he integrates from the bottom to the top.
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.
So ignoring the higher order differentials, we still get an exact answer? 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]
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 is not perimeter times thickness; it's perimeter times the length of the side.
ah, that makes sense. but then volume can also be considered area times some funny ratio and not just the vertical height dz
You can express the length of the side in terms of the height. But that will also depend on the angle.
yeah it will. so you've already explained why we can use dz for volume integrals, by ignoring the higher order differentials. 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.
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. Of course you can use trig and get the length in terms of an angle, if you want.