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So I know the length L of the curve $y=\sqrt{R^{2}-x^{2}}$ from $x=0$ to $x=a$ where $|a| < R$ is given by:

$$L= \int_0^a \frac{R}{\sqrt{R^{2}-x^{2}}}dx $$

Now I must set up the arc length integral and simplify it so that it is in the form listed above.

$$L= \int_0^a \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx$$ and $$\frac{dy}{dx}=-\frac{x}{\sqrt{R^{2}-x^{2}}}$$ $$\left(\frac{dy}{dx}\right)^{2}=\frac{x^{2}}{R^{2}-x^{2}}$$ so $$L= \int_0^a \sqrt{1+\frac{x^{2}}{R^{2}-x^{2}}}dx$$

I am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

adnna1354
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2 Answers2

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Ah never mind I solved it, assuming R and x are positive, $$\sqrt{1+\frac{x^{2}}{R^{2}-x^{2}}}=R\sqrt{\frac{1}{R^{2}-x^{2}}}=\frac{R}{\sqrt{R^{2}-x^{2}}}$$

adnna1354
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Assume $R>a$ and $a,R\in\mathbb{R^+}$:

$$\text{L}=\int_{0}^{a}\sqrt{1+\left(\frac{\partial}{\partial x}\left[\sqrt{R^2-x^2}\right]\right)^2}\space\text{d}x=\int_{0}^{a}\sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}\space\text{d}x=$$ $$\int_{0}^{a}\sqrt{1+\frac{x^2}{R^2-x^2}}\space\text{d}x=\int_{0}^{a}\sqrt{\frac{R^2-x^2}{R^2-x^2}+\frac{x^2}{R^2-x^2}}\space\text{d}x=$$ $$\int_{0}^{a}\sqrt{\frac{R^2-x^2+x^2}{R^2-x^2}}\space\text{d}x=\int_{0}^{a}\sqrt{\frac{R^2}{R^2-x^2}}\space\text{d}x=$$ $$\int_{0}^{a}\sqrt{\frac{\frac{R^2}{R^2}}{\frac{R^2}{R^2}-\frac{x^2}{R^2}}}\space\text{d}x=\int_{0}^{a}\frac{1}{\sqrt{1-\frac{x^2}{R^2}}}\space\text{d}x=$$

Substitute $u=\frac{x}{R}$ and $\text{d}u=\frac{1}{R}\space\text{d}x$.

This gives a new lower bound $u=\frac{0}{R}=0$ and upper bound $u=\frac{a}{R}$:

$$R\int_{0}^{\frac{a}{R}}\frac{1}{\sqrt{1-u^2}}\space\text{d}u=R\left[\arcsin(u)\right]_{0}^{\frac{a}{R}}=R\left(\arcsin\left(\frac{a}{R}\right)-\arcsin(0)\right)=R\arcsin\left(\frac{a}{R}\right)$$

Jan Eerland
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