The surface area of the torus, with the generating circle having radius $r$, and ring “radius” $R$ (measured from the centre of the torus to the centre of the generating circle), is $A=(2\pi r)(2\pi R)=4\pi^{2}rR$. We used here the obvious fact that the centroid of a circle is at its centre.

The volume of the (solid) torus, with the same parameters as above, is $V=(\pi r^{2})(2\pi R)=2\pi^{2}r^{2}R$.

We compute the volume of the three-dimensional ball in $\mathbb{R}^{3}$. The ball can be considered to be the solid of revolution generated by a half-disk. So we will need to know the centroid of the upper-half disk $B^{2}_{+}(r)$, radius $r$, in the plane. By symmetry, this centroid has only a vertical component and no horizontal component. The vertical component is calculated by:

Then the volume of the ball of radius $r$ is

Similarly we can compute the surface area of the sphere of radius $r$, generated by revolving a half-circular arc. The centroid of the upper half-circle $S^{1}_{+}(r)$ in the plane only has the vertical component:

Thus the surface area of the sphere is given by