Finding Moment of Inertia for Rotated Solid: y=sinx, x-axis, [0,pi]

[devious_]

How do I find the moment of inertia of a solid formed by rotating the curve of y=sinx about the x-axis in the interval [0, pi]?

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

 

[Curious3141]

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

 

[devious_]

Yeah, that worked!

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.