# Homework Help: Calculate volume of body T between planes

1. Oct 24, 2013

### skrat

1. The problem statement, all variables and given/known data
Calculate the volume of a body given by plane $z=0$ and $z=1$ and $x^2+y^2+z^2=4$.

2. Relevant equations
$detJ=r^2sin \theta$ for spherical coordinates

3. The attempt at a solution
$V=\iiint_{T}^{}dV=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dz\int_{\theta _{0}}^{\pi /2}r^2sin\theta d\theta$
$V=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dzr^2\frac{1}{2}$
$V=2\pi r^2\frac{1}{2}=\pi r^2=4\pi$

Why is this wroooong? -.-

one more thing: trying to integrate by dx,dy and dz (having some troubles with integral borders):
$V=\int_{0}^{1}dz\int_{-2}^{2}dy\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}}dx$ or not?

Thanks for all the help!

2. Oct 24, 2013

### haruspex

You seem to have a mix spherical and cylindrical coordinates. How are you defining r here? θ should not be going all way to π/2.

3. Oct 24, 2013

### skrat

am... the way I imagined this is that r=2 is constant and $\theta$ angle between R and axis z like on this picture: http://phobos.ramapo.edu/~aferry/wk1principles/principle_images/form.png [Broken]

Hmm, I guess what I need is r as function of z, where r would be radius of a circle at given z.
So...if I am not wrong, this should than work perfectly?

$V=\iiint_{T}^{}dV=\int_{0}^{2\pi }d\varphi \int_{0}^{1}dz\int_{0}^{\sqrt{R_0^{2}+z^2}}rdr$

Where $R_0=2$ is sphere radius.

And if I a not wrong these are cylindrical coordinates?

Last edited by a moderator: May 6, 2017
4. Oct 24, 2013

### skrat

Is it even possible to write this integral with spherical coordinates in one piece or only using the sum of two integrals?

5. Oct 24, 2013

### haruspex

Yes, cylindrical. But you have a sign wrong in the upper bound for r.
You could write it in one piece, but the limits would include functions like min{}, so it's effectively a sum or difference of two.

Last edited by a moderator: May 6, 2017
6. Oct 24, 2013

### jackmell

You could just use volume by washers up the z-axis:

$$V=\int_0^1 \pi r^2 dz$$

and the radius, $r=f(z)$ since it's spherically-symmetric, is only a function of z.

7. Oct 25, 2013

### skrat

I have a similar problem again so I won't start new topic...
Calculate the gravitational force between body S, which is defined as space between $x^2+y^2\leq R$ and $0\leq z \leq R$, and point $T(0,0,h)$ where $h>R$.

2. Relevant equations
$F=G\frac{mM}{r^2}$ where G is constant and r is the distance between tho bodies.

3. The attempt at a solution
Well, my idea was to calculate the gravitational force for a disc with radius R and for distance d above the disk and than sum (integrate) the discs from 0 to z.
Let's try:
$dF=G\frac{mdM}{r^2}$ where m is mass the point and M mass of the cylinder below the point, now here comes the very first problem because $dM=\rho dV$ but $V(r,z)$ so $dM=\rho 2\pi rdrz + \rho \pi r^2dz$

Now what? :/

8. Oct 25, 2013

### skrat

Or I could try

$dF=G\frac{mdM}{d^2}=G\frac{m\rho dV}{d^2}=Gm\rho\frac{dV}{d^2}=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{r}{d^2}dv$ Where dv is thickness of the cylinder.
I also have to take only the vertical component of the force since horizontal component is 0 due to symmetry, thereby
$F=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{rv}{d^3}dv$ where $d=\sqrt{v^2+r^2}$
so

$F=Gm\rho\int_{0}^{2\pi }d\varphi \int_{0}^{R}dr\int_{h}^{h-z}\frac{rv}{(v^2+r^2)^{3/2}}dv$ which now (If I calculated everything right) leads me to some suspicious logarithms:
$F=Gm\rho \pi \begin{bmatrix} ln(1+\frac{R^2}{(h-z)^2})-ln(1+\frac{R^2}{h^2}) \end{bmatrix}$

hmm, if $z=0$ the logarithms are the same therefore the force would be zero which is not what I expected... I more or less expected the force between a disk and point T for z=0...

9. Oct 25, 2013

### haruspex

You've lost me. I thought this was for a disc, so how does the mass of the cylinder come into it? If you meant to write 'mass of the disc', you can't reduce the disc to be equivalent to a point mass at its centre.

10. Oct 25, 2013

### skrat

Well m would be the be the mass of a body in $T$ while M is the mass of the entire cylinder. My idea was to sum up the gravitational force $dF$ in point $T$ which is $v$ above the disc, for all the discs with height $dz$. All the discs together are than whole cylinder from $0$ to $z$.

So yes, what I meant was mass of the disc.

11. Oct 25, 2013

### haruspex

Then you need to integrate over the area of the disc. You don't need to consider a test mass at (0, 0, h), just ask what the field is at that point. What is the field at Cartesian (0, 0, h) due to a disc element of mass $\rho rdr d\theta$ at polar (r, θ) in the XY plane? What component of that field will be relevant to the integration?

12. Oct 25, 2013

### skrat

Am...
$F=Gm\int_{0}^{2\pi }d\theta \int_{0}^{R}\frac{r}{d^2}cos\varphi dr$
$F=Gmh\int_{0}^{2\pi }d\theta \int_{0}^{R}\frac{r}{(h^2+r^2)^{3/2}}$
$F=Gmh2\pi \int_{h^2}^{h^2+R^2}\frac{dt}{t^{3/2}}$
$F=-Gmh\pi \begin{bmatrix} \frac{1}{\sqrt{h^2+R^2}}-\frac{1}{h}\end{bmatrix}=Gmh\pi \begin{bmatrix} \frac{1}{h}-\frac{1}{\sqrt{h^2+R^2}}\end{bmatrix}$

right?

13. Oct 25, 2013

### haruspex

I think you have some factor of two errors. When substituting t, what is dt equal to?
When you then integrated, looks like you multiplied by the -1/2 instead of dividing by it.

14. Oct 25, 2013

### skrat

Ah yes, that's right.

$dt=2rdr$ and than when integrating multiplying by -2.
All together now
$F=Gmh2\pi \begin{bmatrix} \frac{1}{h}-\frac{1}{\sqrt{h^2+R^2}}\end{bmatrix}$ Sorry, I have exactly this on the paper yet...

Now I have to integrate this again, for discs form 0 to z. How do I do that?

15. Oct 25, 2013

### haruspex

First, you need to replace h with a function of z, right?

16. Oct 26, 2013

### skrat

THIS is the step I just couldn't remember!

$F=Gm2\pi \rho (1-\frac{h}{\sqrt{h^2+R^2}})$ where now $h=h_0-z$ therfore $dh=-dz$
$F=-Gm2\pi \rho \int_{0}^{z}(1-\frac{h_0-z}{\sqrt{(h_0-z)^2+R^2}})dz$
$F=-Gm2\pi \rho \begin{bmatrix} \int_{0}^{z}dz-\int_{0}^{z}\frac{h_0-z}{\sqrt{(h_0-z)^2+R^2}}dz \end{bmatrix}$
$F=-Gm2\pi \rho \begin{bmatrix} z+\frac{1}{2}\int_{h_0^2+R^2}^{(h_0-z)^2+R^2}t^{-1/2}dt \end{bmatrix}$
$F=-Gm2\pi \rho \begin{bmatrix} z+\sqrt{(h_0-z)^2+R^2}-\sqrt{h_0^2+R^2} \end{bmatrix}$
or
$F=Gm2\pi \rho \begin{bmatrix} -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2} \end{bmatrix}$

Now second part is $\lim_{R->0}F(R)=\lim_{R->0}Gm2\pi \rho \begin{bmatrix} -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2} \end{bmatrix}$
$\lim_{R->0}F(R)=Gm2\pi \rho \begin{bmatrix} -z-\sqrt{(h_0-z)^2}+\sqrt{h_0^2} \end{bmatrix}=0$

Again, not something I expected. What I would expect is the same force we get over a "stick" Since my english is not very good I tried to draw what I expected (in attachment).
In that case F should be $F=Gm\rho (\frac{1}{h-z}-\frac{1}{h})$

or....not?

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17. Oct 26, 2013

### haruspex

The ρ before you take the limit is a volume density. Taking the limit turns the total mass to 0. Replace ρ with M/πR2 first.

18. Oct 26, 2013

### skrat

You mean $\rho = \frac{M}{\pi R^2z}$ since we are talking about a cylinder here. Right?

$\lim_{R->0}Gm2\pi \rho \begin{bmatrix} -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2} \end{bmatrix}$
$\lim_{R->0}\frac{2GmM}{zR^2}\begin{bmatrix} -z-\sqrt{(h_0-z)^2+R^2}+\sqrt{h_0^2+R^2} \end{bmatrix}=\lim_{R->0}\frac{2GmM}{z}\begin{bmatrix} -\frac{z}{R^2}-\frac{\sqrt{(h_0-z)^2+R^2}}{R^2}+\frac{\sqrt{h_0^2+R^2}}{R^2} \end{bmatrix}$

ammm... Now if R -> infinity, the limit is 0 which is perfect, yet if R -> 0... than... Am? I kind of forgot how to deal with limits such as this one (konst/0)... L hospital doesn't help much.

19. Oct 26, 2013

### haruspex

You need to expand the square roots using the Taylor expansion/binomial theorem for small R.