How to find the centroid of a plane region
Formulas for the centroid
The centroid of a plane region is the center point of the region over the interval ???[a,b]???. In order to calculate the coordinates of the centroid, we’ll need to calculate the area of the region first.
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The equation for this area is
???A=\int^b_af(x)\ dx???
To calculate the coordinates of the centroid ???(\overline{x},\overline{y})???, we’ll use
???\overline{x}=\frac{1}{A}\int^b_axf(x)\ dx???
???\overline{y}=\frac{1}{A}\int^b_a\frac12\left[f(x)\right]^2\ dx???
What is the centroid, and how do we find its coordinates?
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Finding the centroid of a region bounded by specific curves
Example
Find the centroid of the region bounded by the curves ???x=1???, ???x=6???, ???y=0???, and ???y=4???.
First we’ll find the area of the region using
???A=\int^b_af(x)\ dx???
We can use the ???x???-values as the boundaries of the interval, so ???[a,b]??? is ???[1,6]???. To find ???f(x)???, we need to remember that taking the integral of a function is the same thing as finding the area underneath the function. Which means we treat this like an “area between curves” problem, and we get
???A=\int^6_14-0\ dx???
???A=4\int^6_1dx???
???A=4x\big|^6_1???
???A=4(6)-4(1)???
???A=20???
In order to calculate the coordinates of the centroid, we’ll need to calculate the area of the region first.
Now we can use the formulas for ???\bar{x}??? and ???\bar{y}??? to find the coordinates of the centroid. First, let’s solve for ???\bar{x}???.
???\overline{x}=\frac{1}{A}\int^b_axf(x)\ dx???
???\overline{x}=\frac{1}{20}\int^b_ax(4-0)\ dx???
???\overline{x}=\frac{1}{5}\int^6_1x\ dx???
???\overline{x}=\frac15\left(\frac{x^2}{2}\right)\bigg|^6_1???
???\overline{x}=\frac{x^2}{10}\bigg|^6_1???
???\overline{x}=\frac{(6)^2}{10}-\frac{(1)^2}{10}???
???\overline{x}=\frac72???
Now we’ll find ???\overline{y}???.
???\overline{y}=\frac{1}{A}\int^b_a\frac12\left[f(x)\right]^2\ dx???
???\overline{y}=\frac{1}{20}\int^b_a\frac12(4-0)^2\ dx???
???\overline{y}=\frac25\int^6_1dx???
???\overline{y}=\frac{2x}{5}\bigg|^6_1???
???\overline{y}=\frac{2(6)}{5}-\frac{2(1)}{5}???
???\overline{y}=2???
The centroid of the region is at the point ???\left(\frac{7}{2},2\right)???.
