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We compute the length of the piece of the circle between $0$ and $\theta$ for $\theta < \frac{\pi}{2}$ by considering it as the graph of the function $g(y)=\sqrt{1 - y^2}$ as $y$ varies between $0$ and $\sin \theta$.

I am not sure how to interpret the following identity or why it is even true. I understand that we are using arc length, however , not why this integral is equal to theta.

Then $\theta= \int_{0}^{\sin \theta} \sqrt{1 + g'(y)^2}$.

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$g'(y) = \dfrac{-y}{\sqrt{1 - y^2}} \Rightarrow 1 + (g'(y))^2 = 1 + \dfrac{y^2}{1 - y^2} = \dfrac{1}{1 - y^2}$.

Then $\sqrt{1 + (g'(y))^2} = \dfrac{1}{\sqrt{1 - y^2}}$.

The integral of this is, of course, $\sin^{-1} y$, and when you substitute the limits, you get $\theta$. I'm not sure whether this is what you're asking, but you did specifically ask why the integral is equal to $\theta$, even though you know how the integral is the arc length.

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