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.