# How do you find the centroid of the quarter circle of radius 1 with center at the origin lying in the first quadrant?

Mar 11, 2015

Non-Calculus Solution:

Observation 1:
The centroid must lie along the line $y = x$ (otherwise the straight line running through $\left(0 , 0\right)$ and the centroid would be to "heavy" on one side). Observation 2:
For some constant, $c$, the centroid must lie along the line
$x + y = c$ and furthermore, $c$ must be less than $1$ since the area of the triangle formed by the X-axis, Y-axis and $x + y = 1$ is more than half of the area of the quarter circle.

Observation 3:
Since the area of the quarter circle (with radius = $1$ is $\frac{\pi}{4}$
the line $x + y = c$ must divide the quarter circle into $2$ pieces each with area $\frac{\pi}{8}$.

The area of the triangle formed by the X-axis, the Y-axis, and $x + y = c$
is $\frac{{c}^{2}}{2}$

Therefore
$\frac{{c}^{2}}{2} = \frac{\pi}{8}$
$\rightarrow c = \frac{\sqrt{\pi}}{2}$

and the centroid is located at the midpoint of the line segment
$\left(\frac{\sqrt{\pi}}{4} , \frac{\sqrt{\pi}}{4}\right)$