Inertia tensor around principal Axes part 2

Lambda96
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Homework Statement
I need to calculate the inertia tensor again around a principal axes showing the following ##I_{23}=0## and ##I_{22}=I_{33}##.
Relevant Equations
none
Hi,

it's about the task e)

Bildschirmfoto 2022-12-12 um 15.25.21.png

Since the density is homogeneous, I have assumed the following for ##\rho=\frac{M}{V}##.

I then started the proof of ##I_{23}##, the integral looks like this:

$$ I_{23}=\int_{}^{} -\frac{M}{V}r'_2r'_3 d^3r$$

Now I apply the transformation

$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(r_2cos\theta+r_3sin\theta)\Bigr)\cdot \Bigl(-r_2sin\theta + r_3cos\theta \Bigr) \ d^3r$$
$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d^3r$$

Now I just used the clue, so ##\frac{1}{2\pi} \int_{0}^{2\pi} I_{23} d\theta ##

$$I_{23}=\int_{}^{} \frac{1}{2\pi} \int_{0}^{2\pi} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d\theta d^3r $$
$$ I_{23}=\int_{}^{} 0 d^3r=0$$With ##I_{22}=I_{33}## I proceeded as follows

$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_3^2 d^3r $$
$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_2^2 d^3r $$

Then I did the transformation,

$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2-r_2^2sin^2\theta-2r_2r_3sin\theta cos\theta + r_3^2cos^2\theta\Bigr) d^3r $$
$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2+r_2^2cos^2\theta+2r_2r_3sin\theta cos\theta + r_3^2sin^2\theta\Bigr) d^3r $$

Unfortunately, I am now stuck on how to show that the two terms in the integral are equal.
 
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Use <br /> \begin{split}<br /> \cos^2 \theta &amp;= \tfrac12(1 + \cos 2\theta) \\<br /> \sin^2 \theta &amp;= \tfrac12(1 - \cos 2 \theta) \\<br /> \cos \theta \sin \theta &amp;= \tfrac12 \sin 2\theta \\<br /> \end{split}<br /> and recall that the average of sin or cos over a period is zero.
 
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Likes Lambda96 and vanhees71
Thanks pasmith for your help 👍, I was now able to show that ##I_{22}=I_{33}##.
 
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