# 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:

• The body formed will be a half sphere (I assume one edge of the sector is lying along the x axis). This will have a centre of gravity on the x axis at (X, 0) and X can be calculated as an integral. Looking again, maybe you mean $\frac{\pi}{4}$?
– Paul
Commented Mar 2, 2015 at 13:52
• @Paul: I believe the circular sector is given by $$\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 1,x\geq 0, -x\leq y\leq x\}.$$ Commented Mar 2, 2015 at 14:06
• @Paul No not a half sphere, but like a droplet. The angle between the x-axis and either side of the figure is $\pi/4$. Commented Mar 2, 2015 at 14:11
• Is there a name to this (along axis of symmetry stable configuration ) cone + spherical cap? Commented Mar 2, 2015 at 14:25
• OK yes, two integrals then.
– Paul
Commented Mar 2, 2015 at 14:57

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:
$$x_R = \frac{3}{16}\cdot\frac{1}{1-\frac{1}{\sqrt{2}}}=\color{red}{\frac{3}{16}(2+\sqrt{2})}=0.640165\ldots.$$
• Key says the answer is $x=\frac{3R}{16}\frac{1}{1-\frac{1}{\sqrt{2}}}$ Commented Mar 2, 2015 at 14:23