Question 1085863
<font color="black" face="times" size="3">Let
f(x) = 6*sin(2x)
g(x) = 6*cos(2x)
Here is the graph of f(x) and g(x). They intersect at point A. The x coordinate of point A is pi/8 = 0.39 approximately
<img src = "https://i.imgur.com/cMNMouL.png">
The region between the curves, from x = 0 to x = pi/8, is shown by the light blue shading
<img src = "https://i.imgur.com/KTAfuK1.png">
Note: The fact that f(pi/8) = g(pi/8) = 3*sqrt(2) indicates that we would have a fully enclosed region without the need for the right boundary of x = pi/8, so it's a bit redundant. 
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The formulas we'll use can be found <a href="http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx">here</a>. Scroll down til you reach the "Center of Mass Coordinates" section. The formulas in the blue box below are
*[Tex \Large \bar{x} = \frac{1}{A}\int_{a}^{b}x*(f(x)-g(x))dx]
*[Tex \Large \bar{y} = \frac{1}{A}\int_{a}^{b}\frac{1}{2}\left((f(x))^2-(g(x))^2\right)dx]
which represent the coordinates of the centroid. The value of A is the area between the two curves, so,
*[Tex \Large A = \int_{a}^{b}(f(x)-g(x))dx]
Because the red g(x) curve is above the green f(x) curve all throughout the interval 0 < x < pi/8, this means that we must swap the locations and f(x) and g(x) when we subtract. So we should have these three formulas instead
*[Tex \Large \bar{x} = \frac{1}{A}\int_{a}^{b}x*(g(x)-f(x))dx]
*[Tex \Large \bar{y} = \frac{1}{A}\int_{a}^{b}\frac{1}{2}\left((g(x))^2-(f(x))^2\right)dx]
*[Tex \Large A = \int_{a}^{b}(g(x)-f(x))dx]
This is to ensure A is positive and the centroid coordinates end up in the right spot.
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We need to find the area A. Using numerical integration, I get
*[Tex \Large A = \int_{a}^{b}(g(x)-f(x))dx]
*[Tex \Large A = \int_{0}^{\pi/8}(6*\cos(2x)-6*\sin(2x))dx]
*[Tex \Large A \approx 1.24264]
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Now use this to find the x coordinate of the centroid (xbar = *[Tex \large \bar{x}])
Again I'll use numerical integration to make things go quicker and more efficient
*[Tex \Large \bar{x} = \frac{1}{A}\int_{a}^{b}x*(g(x)-f(x))dx]
*[Tex \Large \bar{x} \approx \frac{1}{1.24264}\int_{0}^{\pi/8}x*(6*\cos(2x)-6*\sin(2x))dx]
*[Tex \Large \bar{x} \approx 0.13365]
Do the same for the y coordinate of the centroid (ybar = *[Tex \large \bar{y}])
*[Tex \Large \bar{y} = \frac{1}{A}\int_{a}^{b}\frac{1}{2}\left((g(x))^2-(f(x))^2\right)dx]
*[Tex \Large \bar{y} \approx \frac{1}{1.24264}\int_{0}^{\pi/8}\frac{1}{2}\left((6\cos(2x))^2-(6\sin(2x))^2\right)dx]
*[Tex \Large \bar{y} \approx 3.62132]
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We found that *[Tex \Large \left(\bar{x}, \bar{y}\right) \approx \left(0.13365, 3.62132\right)]
Therefore, the centroid's location is approximately <font color=red>(0.13365, 3.62132)</font>
Here is an updated graph with the centroid point C added in
<img src = "https://i.imgur.com/I9iIk5Q.png">
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