0
$\begingroup$

So I've been tackling the rather nasty integral of...

$\int^R_s\frac{2r}{\sqrt{r^2-s^2}}.\frac{1}{2}(R-r)^2dr$

...where R and s are constants.

However, every method I try I seem to get stumped by something too awkward to integrate (for me at least).

My current method has been by integration by parts.

$u = \frac{1}{2}(R-r)^2, dv = \frac{2r}{\sqrt{r^2-s^2}} $

$du = -(R-r)dr, v = 2\sqrt{r^2-s^2}$

(I got v by using substition of $p = r^2-s^2, dp = 2r.dr$)

So integration by parts, $[uv]^R_s - \int^R_s v.du$

$uv = (R-r)^2\sqrt{r^2-s^2}$

$\int^R_sv.du = -2\int^R_s(R-r)\sqrt{r^2-s^2}dr = 2\int^R_sr\sqrt{r^2-s^2}dr-2R\int^R_s\sqrt{r^2-s^2}dr$

If we take the first integral...

$2\int^R_sr\sqrt{r^2-s^2}dr$, Let $p = r^2-s^2, dp = 2rdr$

=$\int^R_s\sqrt{p.}dp$

= $[\frac{2u^{\frac{3}{2}}}{3}]^R_s$

= $[\frac{2}{3}(r^2-s^2)^{\frac{3}{2}}]^R_s$

Now the second integral is the part I'm stuck on...

$2R\int^R_s\sqrt{r^2-s^2}dr$

I've tried various substitutions, messing around with sec and tan, and I just can't crack it.

Does anyone have any pointers? (Considering what I've done up to this point is correct, I just need this last integral before I can plob my results back into the integration by parts forumla and work out the definite integrals, etc.)

Alternatively, does anyone have suggestions for how else to go about this problem?

EDIT: I have tried substituting $r = sSecθ$, and came to the integral of $2Rs^2\int secθtan^2θdθ$

I'm having trouble integrating this too, as I keep ending back at the same integral.

$\endgroup$
  • $\begingroup$ $r=s\sec\theta$ or $r=s\cosh t$ (and there are others). $\endgroup$ – André Nicolas Mar 10 '14 at 18:32
  • $\begingroup$ @AndréNicolas I've tried sSecθ but I got lost with it. But I imagine it might lead to the right answer. As for hyperbolic functions, we haven't even touched on those before, so it's not something I can really work with at the moment. $\endgroup$ – Wolff Mar 10 '14 at 18:34
  • $\begingroup$ OK, let's use $\sec$, either immediately or after your integrations (which I have not checked). When we substitute, we get, apart from constants, $\int \sec \theta \tan^2\theta\,d\theta$. Equivalently, we want to integrate $\sec\theta-\sec^3\theta$. This is unpleasant. There are "magic" ways, or else we can write $\sec\theta$ as $\frac{\cos\theta}{\cos^2\theta}=\frac{\cos\theta}{1-\sin^2\theta}$, then let $u=\sin\theta$ and use partial fractions. Easier is to look up or remember the integral of $\sec\theta$, or even of $\sqrt{w^2-1}$. $\endgroup$ – André Nicolas Mar 10 '14 at 18:49
1
$\begingroup$

By expanding the $(R-r)^2$ piece, you see that you basically are looking to evaluate

$$\int_s^R dr \frac{r^k}{\sqrt{r^2-s^2}}$$

for $k=1,2,3$. For $k=1$, the integral is simple:

$$\int_s^R dr \frac{r}{\sqrt{r^2-s^2}} = \frac12 \int_{s^2}^{R^2} \frac{du}{\sqrt{u-s^2}} = \sqrt{R^2-s^2}$$

For $k=2$, integration by parts is useful:

$$\begin{align}\int_s^R dr \frac{r^2}{\sqrt{r^2-s^2}} &= \left [r \sqrt{r^2-s^2} \right ]_s^R - \int_s^R dr \, \sqrt{r^2-s^2}\\ &= R\sqrt{R^2-s^2} - s^2 \int_0^{\cosh^{-1}{R/s}}dt \, \sinh^2{t}\\ &= R\sqrt{R^2-s^2} -\frac12 s^2\int_0^{\cosh^{-1}{R/s}}dt \,(\cosh{2 t}-1)\\ &= \frac12 R\sqrt{R^2-s^2} + \frac12 s^2\cosh^{-1}{\frac{R}{s}}\\ &= \frac12 R\sqrt{R^2-s^2} +\frac12 s^2\log{\left (\frac{R}{s} + \sqrt{\frac{R^2}{s^2}-1} \right )} \end{align}$$

For $k=3$, sub and integration by parts again:

$$\begin{align}\int_s^R dr \frac{r^3}{\sqrt{r^2-s^2}} &= \frac12 \int_{s^2}^{R^2} du \frac{u}{\sqrt{u-s^2}}\\ &= \frac12 \left [u \sqrt{u-s^2} \right ]_{s^2}^{R^2} - \frac12 \int_{s^2}^{R^2} du \, \sqrt{u-s^2}\\ &= \frac12 R^2\sqrt{R^2-s^2} - \frac13 \left (R^2-s^2 \right )^{3/2}\end{align}$$

Now put these pieces altogether from the original integral and, voila!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.