Center of gravity of volume formed by rotating a circular sector A circular sector with the radius R and the opening angle $\pi/2$ rotates around its axis of symmetry (the x-axis). A homogenous body is formed. Determine the position of the center of gravity to this body. 
I don't know how to account for the mass of the volume between $R/\sqrt{2} \leq x \leq R$, that is the rounded part of the figure. Any ideas?
Sketch of the figure:

 A: Assuming $R=1$, the circular sector is given by:
$$S=\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 1,x\geq 0, -x\leq y\leq x\}.$$
By splitting the circular sector in many equal thin triangles between two rays emanating from the origin, we easily have that the center of mass of $S$ is the center of mass of the circular arc:
$$\Gamma=\{(x,y)\in\mathbb{R}^2, x^2+y^2=\frac{4}{9}, x\geq 0, -x\leq y\leq x\}.$$
Such center of mass obviously lies on the $x$-axis, and its $x$-coordinate is given by:
$$x_G = \frac{2}{\pi}\int_{-\pi/4}^{\pi/4}\frac{2}{3}\cos\theta\,d\theta={\frac{4\sqrt{2}}{3\pi}}=0.60021\ldots$$
The center of mass of the solid $R$ given by $S$ revolved around the $x$-axis can be computed through Pappus centroid theorem by computing first the volume of the solid $R$ given by $S$ revolved around the $y$-axis, then the area of $S$. The $x$-coordinate of the centroid of $R$ is hence given by:
$$x_R = \frac{3}{16}\cdot\frac{1}{1-\frac{1}{\sqrt{2}}}=\color{red}{\frac{3}{16}(2+\sqrt{2})}=0.640165\ldots.$$
