Calculate the centroid of a semicircle

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Calculate the centroid of a semicircle.

The Correct Answer and Explanation is:

The centroid (or center of mass) of a semicircle can be determined using integral calculus. For a semicircle, the centroid lies along the central axis of symmetry, which is the vertical line passing through the center of the full circle. For a semicircle with radius RRR, we will calculate the distance of the centroid from the flat edge of the semicircle, which lies along the x-axis.

Step-by-step Explanation:

  1. Equation of the Semicircle:
    A semicircle with radius RRR can be represented by the equation: y=R2−x2y = \sqrt{R^2 – x^2}y=R2−x2​ where −R≤x≤R-R \leq x \leq R−R≤x≤R, and the semicircle is above the x-axis.
  2. Finding the Centroid:
    The centroid of a region is given by the formula: yˉ=∫−RRy dA∫−RRdA\bar{y} = \frac{\int_{-R}^{R} y \, dA}{\int_{-R}^{R} dA}yˉ​=∫−RR​dA∫−RR​ydA​ where dAdAdA is the differential area element. For the semicircle, dA=y dxdA = y \, dxdA=ydx, so we substitute this into the formula.
  3. Calculating the Area:
    The total area of the semicircle is: A=∫−RRy dx=∫−RRR2−x2 dxA = \int_{-R}^{R} y \, dx = \int_{-R}^{R} \sqrt{R^2 – x^2} \, dxA=∫−RR​ydx=∫−RR​R2−x2​dx This is a standard integral, and its result is: A=12πR2A = \frac{1}{2} \pi R^2A=21​πR2
  4. Finding the Centroid’s yyy-coordinate:
    Now, we calculate the centroid’s yyy-coordinate: yˉ=∫−RRy2 dx∫−RRy dx\bar{y} = \frac{\int_{-R}^{R} y^2 \, dx}{\int_{-R}^{R} y \, dx}yˉ​=∫−RR​ydx∫−RR​y2dx​ First, calculate the integral of y2y^2y2: ∫−RRy2 dx=∫−RR(R2−x2) dx=πR48\int_{-R}^{R} y^2 \, dx = \int_{-R}^{R} (R^2 – x^2) \, dx = \frac{\pi R^4}{8}∫−RR​y2dx=∫−RR​(R2−x2)dx=8πR4​ Now, divide the integral of y2y^2y2 by the area of the semicircle: yˉ=πR4812πR2=4R3π\bar{y} = \frac{\frac{\pi R^4}{8}}{\frac{1}{2} \pi R^2} = \frac{4R}{3\pi}yˉ​=21​πR28πR4​​=3π4R​

Final Result:

The centroid of a semicircle is located at a distance of 4R3π\frac{4R}{3\pi}3π4R​ from the flat edge along the vertical axis of symmetry. This is approximately 0.424 of the radius from the flat edge.

Thus, the centroid lies 4R3π\frac{4R}{3\pi}3π4R​ units above the flat edge of the semicircle

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