MHB Integral calculus: stellar stereography

[Ciaran]

Hi there, I have a question I'm stuck on. It is:

Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster
from the observed (two-dimensional) density that can be analysed from a photograph. Suppose that in
a spherical cluster of radius R the density of stars depends on ly on the distance r from the centre of
the cluster. If the perceived star density is given by y(s), where s is the observed planar distance from
the centre of the cluster, and x(r) is the actual density, it can be shown that

y(s) = integral from s to R of (2r x(r) dr)/ (r^2-s^2)^1/2 where x(r)=(1/2) ((R-r)^2)

I tried using the trig substitution r= s sec theta but haven't got far. I also am confused about how this substitution changes the limits. Can anyone please help?

 

[MarkFL]

So, we are given:

$$y(s)=\int_s^R \frac{r(R-r)^2}{\sqrt{r^2-s^2}}\,dr$$

The first thing I see is that we have an improper integral, and so I would write:

$$y(s)=\lim_{s\to r}\left[\int_r^R \frac{w(R-w)^2}{\sqrt{w^2-s^2}}\,dw\right]$$

For the integral itself, I would try integration by parts, where:

$$u=(R-w)^2\,\therefore\,du=2(w-R)\,dw$$

$$dv=\frac{w}{\sqrt{w^2-s^2}}\,dw\,\therefore\,v=\sqrt{w^2-s^2}$$

Can you proceed?

 

[Ciaran]

So by integration by parts, we get

(R-w)^2 (w^2-s^2)^0.5- 2 \ int(w-R)*(w^2-s^2)^0.5 dw

 

[MarkFL]

Since it is a definite integral, we would get:

$$\left[(R-w)^2\sqrt{w^2-s^2}\right]_r^R-2\int_r^R(w-R)\sqrt{w^2-s^2}\,dw$$

And this reduces to:

$$(r-R)\sqrt{r^2-s^2}+2R\int_r^R\sqrt{w^2-s^2}\,dw-\int_r^R2w\sqrt{w^2-s^2}\,dw$$

For the first integral, I would suggest the substitution:

$$w=s\coth(u)$$

And for the second, I would try:

$$u=w^2-s^2$$

 

[Ciaran]

So for the first integral, it would be

integral from r to R of (s^2 (sqrt(coth^2u -1))(1-coth^2u) du?

And for the second I get

integral from r to R of 2 sqrt u du

 

[MarkFL]

Let's look first at the integral:

$$\int_r^R\sqrt{w^2-s^2}$$

If we let:

$$w=s\coth(u)$$

Then we find:

$$dw=-s\csch^2(u)\,du$$

And so the integral become:

$$s^2\int_{\arcoth\left(\frac{R}{s}\right)}^{\arcoth\left(\frac{r}{s}\right)}\csch^3(u)\,du$$

I would try playing with identities here to get a form you can integrate directly.

For the second integral, we would get:

$$\int_r^R2w\sqrt{w^2-s^2}\,dw$$

Let:

$$u=w^2-s^2$$

And so:

$$du=2w\,dw$$

And we have:

$$\int_{r^2-s^2}^{R^2-s^2}u^{\frac{1}{2}}\,du$$

 

[Ciaran]

How are you going about changing the limits? I also have never dealt with identities involving hyperbolic cubic functions

 

[Opalg]

Hi there, I have a question I'm stuck on. It is:

$\vdots$

y(s) = integral from s to R of (2r x(r) dr)/ (r^2-s^2)^1/2 where x(r)=(1/2) ((R-r)^2)

I tried using the trig substitution r= s sec theta but haven't got far. I also am confused about how this substitution changes the limits. Can anyone please help?

I like your idea of substituting $r = s\sec\theta$. If you do that, you get $$\begin{aligned} y(s)=\int_s^R \frac{r(R-r)^2}{\sqrt{r^2-s^2}}\,dr &= \int_0^\Theta \frac{s\sec\theta(R - s\sec\theta)^2}{s\tan\theta}\cdot s\sec\theta\tan\theta\,d\theta \\ &= \int_0^\Theta \bigl( sR^2\sec^2\theta -2Rs^2\sec^3\theta + s^3\sec^4\theta\bigr)d\theta, \end{aligned}$$ where $\Theta = \arcsec\frac Rs$ (so that $\tan\Theta = \frac{\sqrt{R^2-s^2}}s$).

If you use the standard integrals $$\int\sec^2\theta\,d\theta = \tan\theta,$$ $$\int\sec^3\theta\,d\theta = \tfrac12\sec\theta\tan\theta + \tfrac12\ln \bigl|\,\sec\theta + \tan\theta\,\bigr|$$ and $$\int\sec^4\theta\,d\theta = \tfrac13\tan^3\theta + \tan\theta,$$ then you should be able to complete the calculation. I make the answer $$y(s) = \tfrac13\sqrt{R^2-s^2}\bigl(R^2 + 2s^2\bigr) - Rs^2\ln\biggl(\dfrac{R + \sqrt{R^2-s^2}}s\biggr).$$

[Mark is absolutely correct that the integral is improper and should really be evaluated as a limit. But in fact the integral converges, and by the time you have made the substitution $r = s\sec\theta$ the singularity has disappeared, because of the way that the $\tan\theta$ in the denominator of the integral cancels with the $\tan\theta$ coming from "$dr = s\sec\theta\tan\theta\,ds$".]

 

[Ciaran]

Hi there,

How come you changed the case of r to R in your expression for theta and the rest of the question but left it in the expression for tan of theta?

 

[Ciaran]

Should it not be lowercase throughout as r= s sec theta implies theta= arcsec (r/s)?

 

[Ciaran]

I get exactly the same except I don't get +2s^2; I get two s^2 terms but one is negative and they end up cancelling.

 

[Opalg]

Should it not be lowercase throughout as r= s sec theta implies theta= arcsec (r/s)?

It is lowercase when it is a variable, as in $r = s\sec\theta$, and it is upper case when it refers to the upper limit of the integral, $R = s\sec\Theta $.

I get exactly the same except I don't get +2s^2; I get two s^2 terms but one is negative and they end up cancelling.

I had $-\frac13s^2 + s^2 = \frac23s^2$. But I haven't been back to re-check it, so I may be wrong.