Oscar Ramos

2023-03-03

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

Hayden Dudley

Beginner2023-03-04Added 6 answers

Non-Calculus Solution:

Observation 1:

The centroid must lie along the line $y=x$ (otherwise the straight line running through (0,0) 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 $\frac{{c}^{2}}{2}$

Thus

$\frac{{c}^{2}}{2}=\frac{\pi}{8}$

$\to c=\frac{\sqrt{\pi}}{2}$

and the centroid is located at the midpoint of the line segment

$(\frac{\sqrt{\pi}}{4},\frac{\sqrt{\pi}}{4})$

Observation 1:

The centroid must lie along the line $y=x$ (otherwise the straight line running through (0,0) 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 $\frac{{c}^{2}}{2}$

Thus

$\frac{{c}^{2}}{2}=\frac{\pi}{8}$

$\to c=\frac{\sqrt{\pi}}{2}$

and the centroid is located at the midpoint of the line segment

$(\frac{\sqrt{\pi}}{4},\frac{\sqrt{\pi}}{4})$