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

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

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Answer 1 · 2015-03-11T18:43:57

Non-Calculus Solution:

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

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

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

Therefore
$\frac{c^{2}}{2} = \frac{π}{8}$ $(c^2)/2 = pi/8$
$\to c = \frac{\sqrt{π}}{2}$ $rarr c = (sqrt(pi))/2$

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

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

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