So I was wondering if anyone can verify this answer. For the

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65KB, 566x480px

Try looking it up on the internet instead of asking, this is trivial and there are myriad resources that go into whatever level of detail you could possibly want.

>>8509878
Wow, calm down, I've tried searching it up, but, there is no concrete answer. The only result I found were questions using this as an integral, without any proof as why r squared is left as a constant. Hence, I'm asking here as a last resort (professor still didn't respond).

>>8509885
r^2 is the jacobian for spherical integrals
google that, then come back to me

>>8509905
From the google searches, Jacobian is multivariable, I haven't done multivariable yet. So I guess, I should just take the general formula as such?

>>8509914
Pretty much. In rough terms, the r^2 is a bookkeeping term that converts spherical to cartesian. Until you've addressed spherical integrals you'll just have to take it as granted.

>>8509924
Thank you, that r^2 has been bugging me for the past few days... finally some relief..

>>8509871
I'm assuming to are talking about [math]I = \int r^{2} dm[/math]

You need to be careful with the notation here. [math]r^{2}[/math] is referring to the distance from the axis of rotation to a mass [math]dm[/math] . Depending on your mass, a change of variables will be needed. Take a uniform solid sphere of radius [math]R[/math] for example. Because the integral is dealing with little bits of mass [math]dm[/math], we need to substitute in a mass density and integrate over that to include familiar coordinates. This will done by
[math]\rho = \frac{dm}{dV} \longrightarrow dm = \rho dV[/math]

where [math]\rho[/math] is the mass density of the sphere, which is constant since we assumed uniformity. Our integral then becomes
[math]I = \int r_{ax}^{2}dm = \int_{V}r_{ax}^{2}dm = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} r_{ax}^{2} \sin^{2}\theta r^{2} dr d\theta d\phi[/math]

Here, I've switched to spherical polar coordinates (google the transformation), which accounts for the change in [math](x,y,z)\rightarrow (r,\theta,\phi)[/math]. If you look up a diagram of spherical coordinates and do some geometry you'll find that [math]r_{ax} = r \sin\theta[/math] meaning [math]r_{ax}^{2} = r^{2}\sin^{2}\theta[/math]. Plugging this into your integral and integrating (look up how to do triple integrals, it's literally just doing three integrals one after the other). You'll get

[math]I = \frac{8}{15}\pi \rho R^{5}[/math]

Note that we still have the mass density. Since we integrated over everything already, we can just use the fact that density = mass/volume, hence [math]\rho = M/ (4/3 \pi R^{2})[/math]. Substitute and reduce to get [math] I = 2/5MR^{2}[/math], which you can confirm in any textbook.

>>8509944
Of course 4chan TeX sucks so let me do some corrections.

I = \int r_{ax}^{2}dm = \int_{V}r_{ax}^{2}dm = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} r_{ax}^{2} \sin^{2}\theta r^{2} dr d\theta d\phi

should be

I = \int r_{ax}^{2}dm = \int_{V}r_{ax}^{2}dV = \int \int \int r_{ax}^{2} dxdydz = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} r_{ax}^{2} \sin^{2}\theta r^{2} dr d\theta d\phi

>>8509947
FUCK ME

Of course 4chan TeX sucks so let me do some corrections.

[math]I = \int r_{ax}^{2}dm = \int_{V}r_{ax}^{2}dm = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} r_{ax}^{2} \sin^{2}\theta r^{2} dr d\theta d\phi[/math]

should be

[math]I = \int r_{ax}^{2}dm = \int_{V}r_{ax}^{2}dV = \int \int \int r_{ax}^{2} dxdydz = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} r_{ax}^{2} \sin^{2}\theta r^{2} dr d\theta d\phi[/math]