#### Answer

$(0,\frac{h}{4})$

#### Work Step by Step

We follow the book's advice to integrate the mass element over dy. Using the volume equation as well as the density, $\rho$, we find:
$M = \int_0^h \frac{\pi r^2 \rho}{h^2} y^2dy$
$M = \frac{\pi r^2 \rho h^3}{3h^2}$
$M = \frac{\pi r^2 \rho h}{3}$
Using the integral equation for the center of mass and plugging in the above value for M, we find that the y center of mass is:
$y_{cm} = \frac{h}{4}$
We know that the shape is symmetrical around the center in terms of x, so we find that the x center of mass is 0.