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In my mind, there are two approaches to finding the average height of a semicircle. Integrating Cartesianally:

$$y_{av}=\frac{1}{2r}\int_{-r}^r\sqrt{r^2-x^2}dx=\frac{\pi r}{4}$$

Or integrating polar:

$$y_{av}=\frac{1}{\pi}\int_{0}^\pi r\sin(\theta)d\theta=\frac{2r}{\pi}$$

Both of these seem intuative as you are integrating the height of all possible inputs, then dividing by the range of inputs to give you the average height of all inputs. Yet, we get different results from each.

Which is the correct average height, and why are they different? Or are they correct in their own manner, and why?

Graviton
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    Briefly, the inputs are weighted differently. – peterwhy Oct 06 '18 at 02:43
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    finding the average height Define what average you want to calculate. First one draws an arbitrary vertical, second one draws a radius at an arbitrary angle. The two define different distributions for the point on the semicircle. – dxiv Oct 06 '18 at 02:45
  • @dxiv, I suppose I'm personally trying to find the height of the "center of mass" of the edge of a semicircle, assuming the thickness can be arbitrarily thin. A.k.a. if the semicircle were to exist, how high up the semicircle would it balance on a horizontal rod perfectly. – Graviton Oct 06 '18 at 02:52
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    For a simple closed curve around the origin, the enclosed area in polar coordinates is given by $\int_{0}^{2\pi}\frac{1}{2}\rho(\theta)^2,d\theta$. If $\rho(\theta)$ constantly equals $r$, such integral is $\pi r^2$ as expected. Also, if you enforce the substitution $x\mapsto r\sin\theta$ in $\frac{1}{r}\int_{0}^{r}\sqrt{r^2-x^2},dx$ you end with $$r\int_{0}^{\pi/2}\cos^2(\theta),d\theta=\frac{\pi r}{4}$$ due to the Jacobian $d\sin\theta=\cos(\theta)d\theta$. Your formula for the area in polar coordinates is essentially assuming that $\sin\theta$ is constant, which is clearly not the case. – Jack D'Aurizio Oct 06 '18 at 02:53
  • @JackD'Aurizio I see, yes it would make a lot more sense to find the equivalent area in polar coordinates, rather than what I had. Thank you. – Graviton Oct 06 '18 at 02:58
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    @Graviton For the centroid of a semicircle arc see 1, 2. For why it is important to fully state what is being asked see Bertrand's paradox. – dxiv Oct 06 '18 at 03:01
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    @dxiv Thank you. Yes, lesson learnt. – Graviton Oct 06 '18 at 03:03
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    I think @Jack is answering a different question about areas enclosed rather than centroids of curves. // For a fun exercise, try applying both your approaches to an L-shaped curve with vertices $(0,1),(0,0),(1,0)$ and it should become clearer what the missing ingredient is. –  Oct 06 '18 at 04:43

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