Moment of Inertia of Thin Plate: x^2+4y^2=4

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SUMMARY

The discussion focuses on calculating the moment of inertia of a thin plate with constant density occupying the region defined by the curve x² + 4y² = 4, specifically about the line x = -3. The integral for the moment of inertia is presented as I(x = -3) = (1/2)ρ∫₀²(x + 3)²√(4 - x²)dx. Participants clarify the correct expression for y(x), confirming that y(x) = (1/2)√(4 - x²) is accurate, addressing previous inconsistencies in the posted information.

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Nyasha
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Consider a thin plate of constant density which occupies the region in the first quadrant inside the curve:

x^2+4y^2=4


Find moment of inertia about line x=-3


Attempt to solution:


y=\frac{\sqrt{4-x^2}}{2}

I(x=-3)=\frac{1\rho}{2}\int_0^2(x+3)^2\sqrt{4-x^2}


\text{Is there any easier way of integrating this thing without having to expand} (x+3)^2 \text{and then multiply it with}\sqrt{4-x^2}
 

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In your solution for y(x), where did the 1/2 come from?

I don't see any options except to do it piece by piece.
 
Dr.D said:
In your solution for y(x), where did the 1/2 come from?

I don't see any options except to do it piece by piece.


y=\sqrt{\frac{4-x^2}{4}}
 
There is a conflict in your posted information. On the figure, it agrees with what you have used
y(x) = (1/2) * sqrt(4-x^2)
I see that now.
In your original post, you also wrote
LaTeX Code: x^2+y^2=4
which leads to
y(x) = sqrt(4-x^2)
That is why I asked the question.
 
Dr.D said:
There is a conflict in your posted information. On the figure, it agrees with what you have used
y(x) = (1/2) * sqrt(4-x^2)
In your original post, you also wrote
LaTeX Code: x^2+y^2=4
which leads to
y(x) = sqrt(4-x^2)
That is why I asked the question.

Thanks for pointing out that mistake :smile:
 

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