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Find the moment of inertia about $x$-axis of the region bounded by $y=x^2$ and $y=x$, if the density is proportional to the distance from the $x$-axis. And the answer should represented by the mass of the region.

I tried first let the density $$\rho=ky$$ Then the moment of inertia is $$\int^1_0\int^x_{x^2}y^2\cdot ky \ dy dx=\frac{k}{45}$$ Then calculate the mass of the region as $$M=\int^1_0\int^x_{x^2}ky \ dydx=\frac{k}{15}$$ So the moment of inertia is $$\frac{M}{3}$$ But the answer is $$\frac{M}{8}$$ There is no detailed solution so I don't know where I got wrong...

Matata
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    I just checked and it looks like your answer is right. The textbook must have gotten it wrong. :) – Tristan Batchler Oct 10 '16 at 03:59
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    @TristanBatchler Thank you for checking~ – Matata Oct 10 '16 at 04:01
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    For what it's worth, I also agree with your answer. I even tried doing the problem for the solid of revolution of that area around the $x$ axis, but it still is not even close to $M/8$. – David K Oct 10 '16 at 04:07

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