- #1

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I've tried to set up integrals by summing up cylinders parallel to the y-axis but to no avail.

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- Thread starter devious_
- Start date

- #1

- 310

- 3

I've tried to set up integrals by summing up cylinders parallel to the y-axis but to no avail.

- #2

Curious3141

Homework Helper

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I'm assuming it's the moment of inertia of the solid about the x-axis that the question is asking for.

Let the volume density of the solid of revolution be [itex]\rho[/itex]. Then a cylindrical element of mass [itex]dm[/itex] is defined by [itex]\rho \pi y^2 dx[/itex]. The moment of inertia of that element about the x-axis is defined by [itex]\frac{1}{2}y^2dm = \frac{1}{2}\rho \pi y^4 dx[/itex]. Substitute [itex]y = \sin x[/itex] and integrate over the required bounds and you have the answer. To remove the [itex]\rho[/itex] term and leave your answer purely in terms of the total mass [itex]M[/itex], just calculate the volume of revolution [itex]V[/itex] the usual way and put [itex]\rho = \frac{M}{V}[/itex].

Let the volume density of the solid of revolution be [itex]\rho[/itex]. Then a cylindrical element of mass [itex]dm[/itex] is defined by [itex]\rho \pi y^2 dx[/itex]. The moment of inertia of that element about the x-axis is defined by [itex]\frac{1}{2}y^2dm = \frac{1}{2}\rho \pi y^4 dx[/itex]. Substitute [itex]y = \sin x[/itex] and integrate over the required bounds and you have the answer. To remove the [itex]\rho[/itex] term and leave your answer purely in terms of the total mass [itex]M[/itex], just calculate the volume of revolution [itex]V[/itex] the usual way and put [itex]\rho = \frac{M}{V}[/itex].

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- #3

- 310

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What I did was EXACTLY the same as your method, except I multipled dm by x^2 instead of y^2. Oops.. :shy:

Thank you.

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