Originally Posted by

**drew.walker** I've been given the following problem to solve:

**The region above the interval [1, 2] of the x-axis and below the curve y = (3x - x^2 - 2)^1/2 is rotated around the x-axis giving a solid of revolution. This solid of revolution has density, depending only on the x-coordinate, this is given by p(x) = 3x. Where is the centre off mass of this solid?**

So far I've worked out the following (which may or may not be correct):

$\displaystyle y = \sqrt{3x - x^2 - 2}$

$\displaystyle A(x) = \pi(\sqrt{3x - x^2 - 2})^2 = \pi(3x - x^2 - 2)$

$\displaystyle Volume = \int^2_1 A(x) dx = \pi \int^2_1 3x - x^2 - 2 dx = \pi \left [ {\frac {3x^2} {2} - \frac {x^3} {3} - 2x} \right ]^2_1$

I think I'm on the right track, but I don't know how I can take this information about the volume and apply it to the density function to produce the mass, then calculate the centre of mass. Any help would be greatly appreciated.