moment of inertia of a rectangular prism

I was trying to derive the formula for the moment of inertia of a rectangular prism at its centre of mass. And ended up with this:

$$\frac{1}{2}M(a/2)^{2}+2\frac{M}{ab}\int^{b/2}_{a/2}arccos(\frac{a^{2}}{2r^{2}}-1)rdr+4\frac{M}{ab}\int^{\sqrt{a^{2}+b^{2}}}_{b/2}(\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr$$

Due to the limitation of my knowledge in Calculus, I was stuck here. Can anyone help me solve these two integral:

$$\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr$$

$$\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr$$
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 Are you trying to calculate one of these?: http://en.wikipedia.org/wiki/List_of_moments_of_inertia Bob S
 Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.

moment of inertia of a rectangular prism

 Quote by benhou Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.
I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

Where ρ is the density. So ρ = mass/xy

You can take it from here.

Bob S
 could you explain how this come about?
 (1/2)*M((1/2)*a)^2+(1/4)*M*(b^4*arccos((2*a^2-b^2)/b^2)*sqrt(-a^2*(a^2-b^2)/b^4)+2*a^4-2*a^2*b^2)/(sqrt(-a^2*(a^2-b^2)/b^4)*b^2*ab)+4*M*(int(((1/2)*Pi-arccos((1/2)*b/r)-arccos((1/2)*a/r))*r, r = (1/2)*b .. sqrt(a^2+b^2)))/ab

 Quote by benhou could you explain how this come about?
I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

Where ρ is the density. So ρ = mass/xy

I =(2m/xy)[ x·y3/(3·8) + y·x3/(3·8)] =(m/12)(y2 + x2)

If x = y, then I = m·x2/6

Compare this to a solid disk of diameter d: I = md2/8

Bob S
 Oh, ya. It's just the idea that we have to use both dx and dy, and integrate them one after the other, also that even though x and y are in different intervals we only let the intervals on the intergral sign worry about it. I just can't grasp it.

 Quote by benhou Oh, ya. It's just the idea that we have to use both dx and dy, and integrate them one after the other, also that even though x and y are in different intervals we only let the intervals on the intergral sign worry about it. I just can't grasp it.
Do the integral for a solid disk of diameter d.
Bob S
 ok, i would start like this: $$\sum r^{2}\Delta m$$ = $$\sum r^{2}\rho A$$ = $$\sum r^{2}\frac{M}{\pi(d/2)^{2}}2\pi r\Delta r$$ = $$8\frac{M}{d^{2}}\sum r^{3}\Delta r$$ = $$8\frac{M}{d^{2}}\int^{d/2}_{0}r^{3}dr$$ = $$8\frac{M}{d^{2}}\frac{(d/2)^{4}}{4}$$ = $$\frac{Md^{2}}{8}$$ Is this how you want me to do it? Or some other way?
 Looks good. Here is a slightly different way: Use r = d/2 and m= pi·d2/4 I = ∫o2pi∫orρ·r2 r·dr dθ I = 2·pi ∫orρ·r2 r·dr I = 2·pi·ρ·r4/4 = pi·ρ·d4/32 = m·d2/8 Bob S
 Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.

 Quote by Bob S ∫-y/2+y/2∫-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy
wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.

 Quote by benhou wait, how did the dy become y and dx become x?
It's a property of definite integrals.Integrating from -y/2 to 0 gives the same result as integrating from 0 to y/2

I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy
 Quote by benhou wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.
Let's take just the first term:

I = ∫-y/2+y/2-x/2+x/2ρ·x2 dx dy

Integrating by y first, then by x:

I = ∫-x/2+x/2 ρ·y·x2 dx = 2∫o+x/2 ρ·y·x2 dx = [2ρ·y/3][x/2]3=2ρ·y·x3/24 = ρ·y·x3/12

Now using ρ = m/xy we get

I = m·x2/12 (just the first term; second term is m·y2/12 )

Bob S
 Cool. Thanks Bob!!!

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