Help! I'm Not Sure What I Did Wrong in Evaluating MOI of a Sphere

[DeadInside]

Homework Statement

Axis is through center of mass of the solid sphere.

Relevant Equations

I first integrated dm to find MOI of a ring.
Then I used MOI of a ring to find MOI of a disc.
Then I used MOI of a disc to find MOI of solid sphere

I know I must have done something wrong somewhere here, but I cannot figure out exactly which one

Answer is supposed to be (2/5)MR²

Whatever disaster I have in the last image does not evaluate closely to that at all.

I'm not looking for another way to find the MOI of solid sphere, I would just like to know what I am doing wrong here. Any input is welcome. Please and thank you.

 

[TSny]

I'm not looking for another way to find the MOI of solid sphere, I would just like to know what I am doing wrong here. Any input is welcome. Please and thank you.

$dh$ should be the thickness of the disc. Draw a figure showing the disc inside the sphere for some angle $\theta$.

 

[DeadInside]

View attachment 287075
$dh$ should be the thickness of the disc. Draw a figure showing the disc inside the sphere for some angle $\theta$.

Hi, thanks for the suggestion. I tried drawing the diagram, but I still don't understand. dh is the thickness already?? I'm really confused.

 

[haruspex]

Hi, thanks for the suggestion. I tried drawing the diagram, but I still don't understand. dh is the thickness already?? I'm really confused.

Your diagram (3rd image in post #1) shows dh as a distance around the circumference of the sphere. Your expression $dh=Rd\theta$ conforms to that. But that is not the thickness of the disc, which is what you need dh to be according to your integral.

 

[Delta2]

There is a much simpler way to calculate this, instead of splitting to $I_{ring}$ and $I_{disc}$ as intermediate steps, to calculate the raw volume integral of MoI in spherical coordinates for the volume of a sphere.
The final integral has the form $$\int_0^{2\pi}\int_0^{\pi}\int_0^R \rho r^4\sin^3\theta dr d\theta d\phi$$. I know you are interested mainly in spotting the flaws in your specific attempt so I won't interfere any further.

 

[DeadInside]

View attachment 287075
$dh$ should be the thickness of the disc. Draw a figure showing the disc inside the sphere for some angle $\theta$.

At first, I couldn't understand, but what you said clicked later on.

Your diagram (3rd image in post #1) shows dh as a distance around the circumference of the sphere. Your expression $dh=Rd\theta$ conforms to that. But that is not the thickness of the disc, which is what you need dh to be according to your integral.

I followed this and revisited the relationship between $dh$ and $d\theta$. But I still was unsure what I had to change.

There is a much simpler way to calculate this, instead of splitting to $I_{ring}$ and $I_{disc}$ as intermediate steps, to calculate the raw volume integral of MoI in spherical coordinates for the volume of a sphere.
The final integral has the form $$\int_0^{2\pi}\int_0^{\pi}\int_0^R \rho r^4\sin^3\theta dr d\theta d\phi$$. I know you are interested mainly in spotting the flaws in your specific attempt so I won't interfere any further.

Although this isn't exactly what I looked for, but it hinted me that I may have ignored that:

1) as TSny said, $dh$ should ONLY be the thickness of the disk and ring. It shouldn't be readily assumed as arc length of a small circular segment. Numerically they might be close, but by principle they are not the same.

2) as haruspex further said, because I assumed $dh=Rd\theta$, it restricts $dh$ to a small arc length, which obviously isn't true when it was first established as thickness of disk and ring.

This all clicked when I saw the solution you offered. I thought there must be a trigonometric relationship between arc length and thickness $dh$.

I don't have access to PC for a while so I will add the solution a few days later.

 

[DeadInside]

View attachment 287075
$dh$ should be the thickness of the disc. Draw a figure showing the disc inside the sphere for some angle $\theta$.

Your diagram (3rd image in post #1) shows dh as a distance around the circumference of the sphere. Your expression $dh=Rd\theta$ conforms to that. But that is not the thickness of the disc, which is what you need dh to be according to your integral.

There is a much simpler way to calculate this, instead of splitting to $I_{ring}$ and $I_{disc}$ as intermediate steps, to calculate the raw volume integral of MoI in spherical coordinates for the volume of a sphere.
The final integral has the form $$\int_0^{2\pi}\int_0^{\pi}\int_0^R \rho r^4\sin^3\theta dr d\theta d\phi$$. I know you are interested mainly in spotting the flaws in your specific attempt so I won't interfere any further.

This is the temporary brief explanation I drew on my phone:

The angle here can be found simply with some subtraction and addition of some polygon angles.

This is the actual relation between $dh$ and $Rd\theta$.

Thank you all brilliant and kind minds so so much for your inputs. I have finally ended my 3 days of agony.

 

[haruspex]

This is the actual relation between dh and Rdθ.

Well done.
You may also be unaware of an interesting bit of geometry related to the surface area.
If you project the disc out to meet the coaxial cylinder enclosing the sphere, the surface area of the sphere within the slice equals the surface area of the cylinder within the slice; both are $2\pi R\sin(\theta)d\theta$. Thus the surface area of the whole sphere is that of the cylinder: $(2\pi R)(2R)=4\pi R^2$. This was first realized by Archimedes, making him effectively the discoverer of calculus.