# Homework Help: Centre of mass/double integral question

1. May 16, 2010

### nugget

1. The problem statement, all variables and given/known data

Find the centre of mass of the 2-dimensional plate which occupies the region inside the circle x2+y2=2y, but outside x2+y2=1, and for which the density is inversely proportional to its distance from the origin.

2. Relevant equations/workings

The first circle can be rearranged to give x2+(y-1)2=1, i.e. a circle of radius 1 with centre at (0,1).

The last piece of information translates to p=1/(x2+y2)1/2, where p = density.

The formula for mass of this plate will be a double integral of p over the given area. Sketching a diagram, we can see that this area is the upper part of the first circle, (i.e. the area of the first circle minus the intersecting area of two circles).

Once we have the mass, centre of mass is discernible via other formulae.

I basically need help to know what my upper and lower values should be to integrate over for x and y, and also how to initially integrate p with respect to y... this should yield a trig or log function, right?

2. May 16, 2010

### nugget

Looking at a plot of the circles, I get the feeling I will need to split the double integral into two parts and then integrate with x as my inner variable over the interval [sqrt(1-y2,sqrt(2y-y2)] and integrate from 0 to 2 in y. (I only need to work out one of the two integrals i get by splitting and can then multiply it by 2 as they are symmetrical)

Does this sound about right? I basically rearranged the 2 circle formulae to make x the subject and then put the lower circle as my lower value on the interval and the higher one as my higher value...

so i end up with:

m = 2*{integral over 0 to 2} {integral over sqrt(1-y2,sqrt(2y-y2)} of 1/(sqrt(x2+y2)) dxdy

Last edited: May 16, 2010
3. May 16, 2010

### nugget

would it be easier/possible if i changed coordinate systems?

4. May 16, 2010

### tiny-tim

Hi nugget!
much easier! …

except, you don't need new coordinates, just one new parameter

use r, and you'll find you only need a single integral (not a double one).

5. May 17, 2010

### nugget

Thanks Tim,

I think I'm correct in saying that the density is now:

P = K/r ----> (K is some constant)

The x-coordinate of my centre of mass should be zero as the lamina is symmetrical about the y-axis.

To find the y-coordinate I must first find the mass of the plate.

Mass is (according to my textbook) equal to a double integral of P over the area of the lamina.

If that's correct, then how will I end up with a single integral?

If I convert to polar coordinates, the dxdy becomes r.dr.d(theta).

Hence the mass equals a double integral of: K dr.d(theta)

If this is correct then that's great. What I don't understand is what theta values I am supposed to integrate over? I assume the r values are 0 and 1 because 1 is the radius of each of my circles.

any clues?

I know that the y coordinate of my centre of mass is equal to:

(1/m)*double integral of x multiplied by p.

so if I convert to polar, then this should be:

(K/m)*double integral of r.cos(theta) dr.d(theta)

Thanks for your help,

p.s. how do I write using real notation in these posts?

Last edited: May 17, 2010
6. May 17, 2010

### tiny-tim

Hi nugget!

(I'm sorry I didn't reply earlier: the PF server seemed to stop working, and then I had to go out )

(have an integral: ∫ and a theta: θ )
For each value of r, your limits for θ will be between the minimum and maximum values of θ for that value of r.

Yes, of course, the integral is ∫∫ … r drdθ, but in practice you can instantly see that (if the density is ρ) the mass of the arc-shaped strip of radius r and thickness dr is simply ρr(θmax(r) - θmin(r))dr.

So you can just do one integration … ∫ ρr(θmax(r) - θmin(r)) dr.

Of course, this is the same as the double integration, ∫∫θmin(r)θmax(r) ρr dθdr, because the first integration gives you [θ]θmin(r)θmax(r), which is θmax(r) - θmin(r).

This cutting-down on double or triple integrations is standard practice … for example, in finding the volume of a volume of revolution, it usually involves dividing the region into disc-shaped or cylindrical slices whose volume is obvious.

Last edited: May 17, 2010
7. May 18, 2010

### nugget

Hi again Tim,

Thanks for your help, although I'm not quite sure that I understood everything you said.

What I took from that post was that my upper and lower limits for θ are going to be π/2 and 0, respectively.

The mass that I end up with is Kπ/2, and the centre of mass is therefore (0,2K/π)

I'm pretty sure that I have answered this question incorrectly but I just didn't grasp what you meant by
If my answers are correct I would still like to know why exactly 0 and π/2 were my interval for θ...

Sorry to be a pain :P Does this question require an understanding of/employ formulae regarding arc length?

Thanks

8. May 18, 2010

### tiny-tim

For each value of r, the curve r = constant is an arc centred on the origin which intersects the upper circle.

I don't understand where you get 0 and π/2 from.

The limits for θ are the values of θ (from the origin) at the ends of this arc (they will be± something).

9. May 18, 2010

### nugget

Hi, are my theta values π/6 and 5π/6?

If so then my mass = 2Kπ/3.

10. May 18, 2010

### tiny-tim

The limits of θ depend on r … they're different for each arc.

11. May 18, 2010

### nugget

The mass I'm after is:

m = K∫∫drdθ. right?

I really don't know where to go anymore; although I am able to visualise the problem I don't get how I am supposed to find the area of the upper circle minus the intersecting area of the two circles...

Do I need to use the fact that my upper circle has equation r = 2sin(θ), and my lower circle has equation r = 1?

12. May 18, 2010

### nugget

...Does my upper circle have this equation? man this is complicated

13. May 18, 2010

### tiny-tim

Yes!

The arc at r = 1.6, say, intersects the upper circle where sinθ = 0.8, ie at θ = sin-10 and π - sin-10.8.

So the limits for the θ integration are ∫12 dr∫sin-1rπ - sin-1r dθ.

Alternatively, you can do the integrals the other way round …

π/65π/6 dθ∫12sinθ dr …

which now I look at it carefully is probably easier to calculate (the first one involves an awkard sin-1 integration )

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