$$\bar{\cos{\theta}} = \int dr \, r^2 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, \cos{\theta}$$
$$\bar{\sin{\phi}} = \int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \, \int d\theta \, $$
$$\bar{r} = \int dr \, r^3 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, $$
Then
$$\begin{align}\bar{r} \,\bar{\sin{\phi}}\,\bar{\cos{\theta}} &= \frac{\int dr \, r^3 f(r)\left (\int dr \, r^3 f(r) \right )^2 \int d\phi \, \sin^2{\phi} \left ( \int d\phi \, \sin{\phi} \right )^2\int d\theta \, \cos{\theta} \left ( \int d\theta \right )^2}{\left (\int dr \, r^2 f(r) \right )^3 \left ( \int d\phi \, \sin{\phi} \right )^3 \left ( \int d\theta \right )^3}\\ &= \frac{\int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \,\, \int d\theta \, \cos{\theta} }{\int dr \, r^2 f(r)\, \int d\phi \, \sin{\phi}\, \int d\theta }\\ &= \bar{x}\end{align}$$
But I do not think you can work with the bare $\phi$ and $\theta$ because the trig functions introduce a nonlinearity that belies the linearity of the integrals.
This is the Pappus-Guldin Theorem.
The volume is the area times the distance travelled by the centroid.
From a calculus point of view, it can be verified by looking at the formula for the volume of the solid of revolution (Method of Shells), and the formula for the ($y$-coordinate of the) centroid.
Best Answer
You can also treat the circular plate as a "composite object" made up of two semi-circular plates of different uniform densities. If you already know (or can calculate easily, or even just look up) that the centroid of a uniform semi-circular "lamina" of radius 1 is $ \ \frac{4}{3 \pi} \ $ above its center, then the centroid of the "upper" semicircle lies at, say, $ \ y \ = \ + \frac{4}{3 \pi} \ $ and the centroid of the "lower" one at $ \ y \ = \ - \frac{4}{3 \pi} \ $ .
The upper plate has twice the mass of the lower one, so the centroid of the entire circle lies two-thirds of the way toward the "upper" centroid along the line joining the two centroids (using a "weighted average" -- in fact, the origin of the term). (This line lies along the "vertical" line of symmetry of the entire circle, so $ \ \overline{x} \ = \ 0 \ $ ) .