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$r' = \sqrt{r^2 + a^2 - 2arcos\theta}$

Why, then, is $2r'dr' = (2r - 2acos\theta)dr$

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Note - cursive $r$ in the diagram is written as $r'$ above.

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    Square the formula & differentiate (assume $a$ & $\theta$ remain constant.) – 2017-02-01

1 Answers 1

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$$r' = \sqrt{r^2 + a^2 - 2arcos\theta}=(r^2 + a^2 - 2arcos\theta)^{1/2}$$

$$dr'=\frac{1}{2}(r^2 + a^2 - 2arcos\theta)^{-1/2}(2r-2a\cos\theta)dr$$

$$dr'=\frac{1}{2}\frac{1}{\sqrt{r^2 + a^2 - 2arcos\theta}}(2r-2a\cos\theta)dr$$

$$dr'=\frac{1}{2}\frac{1}{r'}(2r-2a\cos\theta)dr$$

$$2r'dr'=(2r-2a\cos\theta)dr$$