244. Centers of Gravity. Find the centers of gravity of a homogeneous semicircle and a homogeneous semicircular area.
244. It is evident that the center of gravity of a homogeneous semicircle of radius $r$ lies on the perpendicular to the diameter, erected from the center of the circle, at a distance $y$ from this diameter. If we rotate our semicircle around the diameter, we will obtain a sphere. The first theorem of Pappus states that the surface area formed by the rotation of a plane curve around a straight line, which lies in the same plane and does not intersect the curve, is equal to the product of the length of the curve and the length of the circumference described by its center of gravity. In our case, $4 \pi r^{2}=\pi r(2 \pi y)$, from which $y=\frac{2 r}{\pi}$.

[Clamkin, M. M., 26, 226 (March 1953).]

The second theorem of Pappus states that the volume of the solid formed by the rotation of a plane area around a straight line, which lies in the same plane and does not intersect the area, is equal to the product of the area of the region and the length of the circumference described by its center of gravity.

The center of gravity of a homogeneous semicircle, obviously, lies on the radius $r$, perpendicular to the diameter, at a distance $z$ from this diameter. When the semicircle is rotated around this diameter, we obtain a sphere. Therefore, $\frac{4 \pi r^{3}}{3}=\left(\frac{\pi r^{2}}{2}\right)(2 \pi z)$, so $z=\frac{4 r}{3 \pi}$.

[Clamkin, M. M., 27, 227 (March 1954).]