SOLUTION: what is the centroid of the region bounded by the given curves.
y = 6 sin(2x), y = 6 cos(2x), x = 0, x = π/8
Thanks for your help .
Algebra.Com
Question 1085863: what is the centroid of the region bounded by the given curves.
y = 6 sin(2x), y = 6 cos(2x), x = 0, x = π/8
Thanks for your help .
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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

The region between the curves, from x = 0 to x = pi/8, is shown by the light blue shading

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 here. Scroll down til you reach the "Center of Mass Coordinates" section. The formulas in the blue box below are
-g(x))dx)
)^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,
-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
-f(x))dx)
)^2-(f(x))^2\right)dx)
-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
-f(x))dx)
-6*\sin(2x))dx)

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Now use this to find the x coordinate of the centroid (xbar =
)
Again I'll use numerical integration to make things go quicker and more efficient
-f(x))dx)
-6*\sin(2x))dx)

Do the same for the y coordinate of the centroid (ybar =
)
)^2-(f(x))^2\right)dx)
)^2-(6\sin(2x))^2\right)dx)

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We found that  \approx \left(0.13365, 3.62132\right))
Therefore, the centroid's location is approximately (0.13365, 3.62132)
Here is an updated graph with the centroid point C added in

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