Vector calculus: The distribution of mass on a hemispherical shell

**Jnix** · May 18th 2011, 02:30 AM

Hi all,

This is my first post on MHF. I have been trying to improve my basic calculus skills, working through an (excellent) textbook called "div, grad, curl and all that" by H. M. Schey.

I have just finished reading the second chapter on surface integrals and the divergence and am working on the problems at the end of the chapter. Problem II-4 has me stumped.

It is as follows:

The distribution of mass on the hemispherical shell,
$z = (R^{2}-x^{2}-y^{2})^{1/2}$
is given by
$\sigma (x,y,z) = (\frac{\sigma_{0}}{R^{2}})(x^{2}+y^{2})$
where $\sigma_{0}$ is a constant. Find an expression in terms of $\sigma_{0}$ and R for the total mass of the shell.

I know the solution is $4 \pi R^{2}\frac{\sigma_{0}}{3}$ but am not sure how to get there.

I started off with a surface integral

$\iint_{S} \sigma (x,y,z) dS$

where S is the surface mentioned above.

Now I know to evaluate this, I need to recast the problem in terms of an integral over the projection of the surface on a plane, i.e. the xy-plane. This projected surface (a circle) will be called Q.

I think the first step towards achieving this is the following:

$\iint_{Q} \sigma [x,y,z(x,y)] \left( 1 + \frac{\partial z(x,y)}{\partial x} + \frac{\partial z(x,y)}{\partial y} \right)^{1/2} dQ$

This is where I am stumped. For a start, $\sigma$ doesn't actually seem to be a function of z -- it only depends on x and y. I'm also not sure what role (if any) R plays in this. I assume this is the radius of the hemisphere, but am not sure.

I'd be really grateful if someone could point me in the right direction with what to do next here.

Many thanks!

**Mondreus** · May 18th 2011, 03:26 AM

Your work looks good so far.

Originally Posted by Jnix

This is where I am stumped. For a start, $\sigma$ doesn't actually seem to be a function of z -- it only depends on x and y. I'm also not sure what role (if any) R plays in this. I assume this is the radius of the hemisphere, but am not sure.!

$\sigma$ is a function of x, y and z, but it does not need to depend on z in the same way that $f(x)=1$ does not depend on x.

R is a constant tells us what the radius of the hemisphere is for each $0 \leq z \leq R$

If you continue the same way and switch to polar coordinates with $0 \leq r \leq R$ and $0 \leq \varphi \leq 2\pi$ you should end up with the right answer (it came out right when I did it).

**Jnix** · May 18th 2011, 05:14 AM

Thanks.

So, converting to polar coordinates, we have

$z = (R^{2}-r^{2}\sin^{2}\theta - r^{2}\cos^{2}\theta)^{1/2} = (R^{2}-r^{2})^{1/2}$

since $\sin^{2}\theta+\cos^{2} \theta = 1$ ,

$\sigma (x,y,z) = \frac{\sigma_{0}}{R^{2}} r^{2}$

and

$(1-\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2})^{1/2}=(1-\frac{r^{2}}{(R^{2}-r^{2})^{1/2}})^{1/2}$

using the trigonometric relationship above and the derivatives of z(x,y) wrt x and y: -x/z and -y/z, respectively.

The Jacobian is just $r$ (i.e. dx dy = dr d $\theta$ r).

Now I have:

$\frac{\sigma_{0}}{R^{2}}\int_{0}^{r}\int_{0}^{2\pi } r^{3}\left(1-\frac{r^{2}}{(R^{2}-r^{2})^{1/2}}\right)^{1/2} dr d\theta$

I think I may have gone wrong somewhere! I would have thought that a pen-and-paper integral in this textbook would have a simpler integrand.. Any ideas?

Thanks again for your help.

**Mondreus** · May 18th 2011, 05:29 AM

You've made a slight error. The integral should be $\frac{\sigma_0}{R^2}\iint r^3\sqrt{1-\frac{r^2}{R^2-r^2}}drd\theta = \frac{\sigma_0}{R}\iint \frac{r^3}{\sqrt{R^2-r^2}}drd\theta$

You can verify that you get the right answer here: http://integrals.wolfram.com/index.j...D&random=false

**Mondreus** · May 18th 2011, 08:10 AM

The last integral can be solved like this:
$\int \frac{r^3}{\sqrt{R^2-r^2}}dri=\int -r\left(\frac{R^2-r^2}{\sqrt{R^2-r^2}}-\frac{R^2}{\sqrt{R^2-r^2}}\right)dr=\int \left(-r\sqrt{R^2-r^2}+r\frac{R^2}{\sqrt{R^2-r^2}}\right)dr=\left[\frac{1}{3}(R^2-r^2)^{3/2}-R^2\sqrt{R^2-r^2}\right]=\left[-\frac{1}{3}(2R^2+r^2)\sqrt{R^2-r^2}\right]$

**Jnix** · May 19th 2011, 01:17 AM

Thanks Mondreus.

I understand the steps you outline, but just can't work out how you got from

$\frac{r^{3}}{R}\sqrt{1-\frac{r^{2}}{R^{2}-r^{2}}}$

to

$\frac{r^{3}}{\sqrt{R^{2}-r^{2}}}$ .

There's probably a simple answer, I just seem blind to it.

**Mondreus** · May 19th 2011, 04:16 AM

Sorry, it should be $\sqrt{1+ \frac{r^2}{R^2-r^2}}=\sqrt{1- \frac{-r^2}{R^2-r^2}}=\sqrt{1- \frac{R^2-r^2}{R^2-r^2}+\frac{R^2}{R^2-r^2}}=\sqrt{\frac{R^2}{R^2-r^2}}=\frac{|R|}{\sqrt{R^2-r^2}}$

I worked it out myself, but then copied your surface element (which is wrong) when I posted here.

$dS = \left|\left(\frac{\partial z}{\partial y}\right)^2+\left(\frac{\partial z}{\partial x}\right)^2+1 \right|dxdy=\sqrt{\frac{x^2+y^2}{R^2-x^2-y^2}+1}dxdy$

**Jnix** · May 23rd 2011, 01:45 AM

Originally Posted by Mondreus

The last integral can be solved like this:
$\int \frac{r^3}{\sqrt{R^2-r^2}}dri=\int -r\left(\frac{R^2-r^2}{\sqrt{R^2-r^2}}-\frac{R^2}{\sqrt{R^2-r^2}}\right)dr=\int \left(-r\sqrt{R^2-r^2}+r\frac{R^2}{\sqrt{R^2-r^2}}\right)dr=\left[\frac{1}{3}(R^2-r^2)^{3/2}-R^2\sqrt{R^2-r^2}\right]=\left[-\frac{1}{3}(2R^2+r^2)\sqrt{R^2-r^2}\right]$

Many thanks. I can now get to the final form of the integral,
$\int_{0}^{R}dR \int_{0}^{2\pi} d\theta \frac{r^{3}}{\sqrt{R^{2}-r^{2}}}$

I'm a bit confused as to the logic of your integration, though. My pen and paper integration skills aren't up to much, so if you could just let me know what you are doing at each step (i.e. using inspection, substitution, integration by parts), that'd be a great help to me.

Thanks again!

**HallsofIvy** · May 23rd 2011, 02:21 AM

Originally Posted by Jnix

Many thanks. I can now get to the final form of the integral,
$\int_{0}^{R}dR \int_{0}^{2\pi} d\theta \frac{r^{3}}{\sqrt{R^{2}-r^{2}}}$

I'm a bit confused as to the logic of your integration, though. My pen and paper integration skills aren't up to much, so if you could just let me know what you are doing at each step (i.e. using inspection, substitution, integration by parts), that'd be a great help to me.

Thanks again!

I consider it bad practice to write the integrand "outside" the integral like that. Yes, many texts write " $\int dx f(x)$ " but that can be misleading- especially in double integrals or when integrals appear in equations. In any case, $\int_{r=0}^R\int_{\theta= 0}^{2\pi}\frac{r^3}{\sqrt{R^2- r^2}} d\theta dr= \left(\int_{\theta= 0}^{2\pi} d\theta\right)\left(\int_{r=0}^R\frac{r^3}{\sqrt{R ^2- r^2}}dr\right)$ .

Obviously $\int_{\theta= 0}^{2\pi} d\theta= 2\pi$ so that integral is $2\pi \int_{r= 0}^R \frac{r^3}{\sqrt{R^2- r^2}}dr$ . I would do that integral, equivalently to what Mondreus does, but, I think, logically simpler, by thinking of it as
$2\pi \int_{r= 0}^R \frac{r^2}{\sqrt{R^2- r^2}}(r dr)$

Now, let $u= R^2- r^2$ so that du= -2rdr and $rd= -\frac{1}{2}du$ . Since $u= R^2- x^2$ , $x^2=R^2- u$ and when r= 0, u= R^2, when r= R, u= 0. The integral becomes
$2\pi \int_{u= R^2}^0 \frac{R^2- u}{\sqrt{u}}\left(-\frac{1}{2}du\right)$
$= \pi \int_{u= 0}^{R^2}\frac{R^2- u}{u^{1/2}} du$
where I have used the "-" in $-\frac{1}{2}du$ to swap the limits of integration.

$=\pi R^2\int_{u=0}^{R^2} u^{-1/2}du- \pi\int_{u= 0}^{R^2} u^{1/2} du$

**Jnix** · May 24th 2011, 01:33 AM

Thank you HallsOfIvy, that was a very clear and helpful explanation!

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