Finding the Centroid of a Parabola: Calculating Coordinates for y = 10m

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The equation y = x2 describes a parabola. Find the x and y coordinates of the centroid of the area bounded by this curve, and the line y = 10m
 
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OK, show us what you have tried.
 
I tried to intergrate from -3.16 to 3.16, b/c if you draw a perpinduclar line down from y=10 it intersects the x-axis at those two values. I integrate the given equation y=x^2, but the number I got was 21, so i do not know if it is correct.
 
I think this problem is asking you to determine the centroid of the area below y = 10 m and above y = x^2. Can you calculate the area of the enclosed parabola?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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