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$$\int \left ( r\sqrt{R^2-r^2} \right )dr$$

It looks simple. I know that the derivative of

$$\left (R^2-r^2 \right )^\frac{3}{2}$$

Is the stuff in the integral.

However, what about if I don't know?

How in general do we solve integral of


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try $u=R^2-r^2$ – Yimin Mar 6 '13 at 4:02
you are asking how to find an anti-derivative of an arbitrary function raised to an arbitrary power, am I getting that right? – Jonathan Mar 6 '13 at 4:04
I don't think you can find a general antiderivative of $r \mapsto G^n(r)$. If you have $r \mapsto G^n(r) G'(r)$, as in your example above, then you have a general solution. – copper.hat Mar 6 '13 at 4:43

Note that $(r^2)' = 2r$, so that $(R^2-r^2)' = -2r$ (with respect to $r$).

Therefore, if $u = R^2 - r^2$, $du = -2r\ dr$, so, for any function $f$, $\int r f(R^2-r^2) dr = -\frac1{2}\int (-2r) f(R^2-r^2) dr = -\frac1{2}\int f(R^2-r^2) (-2r\ dr) = -\frac1{2} \int f(u) du $.

Putting $f(u) = \sqrt{u}$ gets the desired result.

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This kind of integral is usually dealt with standard trignometry substitutions. You can use $r=R\cos[\theta]$, $\sqrt{R^{2}-r^{2}}=R\sin[\theta]$, $dr=-R\sin[\theta]d\theta$, for example. I am sure you can find some other ingenius ways to do the job as well.

The main difficulty for this approach is to integrate $\cos[\theta]\sin^{2}[\theta]=\frac{1}{4}\sin[2\theta]\sin[\theta]=\frac{1}{8}(\cos[2\theta-\theta]-\cos[2\theta+\theta])$. You may also try to integrate by parts.

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The OP is asking about $\int r \sqrt{R^2 - r^2} dr$ not $\int \sqrt{R^2 - r^2} dr$. – JavaMan Mar 6 '13 at 4:25

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