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# Triple Integral

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A spherical cloud of gas of radius 3km is more dense at the center that toward the edge, at a distance p km from the center the density is d(p) = 3 - p, write an integral representing the total mass of the cloud and evaluate it.

My problem is not evaluating the integral it is setting it up.

Clearly the best option is a triple integral in spherical coordinates.

/pi  / 2pi / 3
|    |      |        d(p)p^2 sin(x) dp dy dx
|    |      |
/0  /0    /0

Here is my problem, I've tried with d(p) = 3-p like the problem states and I get a very strange answer, the homework solution uses p - 3, and I dont see why that make this change. Please help.

-Brian
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Commented:
If you take the surface area of concentric spheres 4*pi*r^2
The mass of a thin spherical shell of radius r1 and thickness dr is:
(3-r1)4*pi*r1^2 dr

So I think you just need to integrate this from 0 to 3 wrt r.

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Commented:
Hi JR2003,

Thanks for the response. I'm pretty sure that a sitution such as this setup in cartesian would be quite a bit messier.

-Brian

Commented:
disregard last statment.
Commented:
There are a lot of ways to set up the integral.  I think you have the right idea.

And  p-3 is just  -d(p), by the way; if you set the limits up differently, that can turn up.

Whether they picked the same limits matters.

In what way is that a strange answer?

Commented:
>>In what way is that a strange answer?

It is completly wrong. Should I post my complete work?
Commented:
Doing a spherical integral is just way too complex.  Take advange of the simple geometry of the situation.
Surface of a sphere of radius r is 4*pi*r^2
density is 3-r

So, total mass of the cloud is the integral of (3-r)*(4*pi*r^2)dr from 0 to 3.

(3-r)*(4*pi*r^2)dr
(12*pi*r^2 - 4*pi*r^3)dr
This integrates to:
4*pi*r^3 - pi*r^4
Evaluate this from 0 to 3
4*pi*27 - pi*81 - (0 - 0)
27pi

Commented:
>>Doing a spherical integral is just way too complex.

I have to do it in either cylindrical or spherical coordinates, for this section i cannot use cartesian coordinates.
Commented:
Not complicated at all.

G(x,y,p) =  d(p)p^2 sin(x)
== ( 3p^2 - p^3 ) sin(x)

Integrate[ G(x,y,p),  {p from 0 to 3} ]  =>
( p^3 - p^4/4 ) sin(x)  |{p from 0 to 3}   =>
H(x) =>  ( 27 -  81/4  ) sin(x)  => 27/4  sin(x)

Integrate [ H(x) ]{y from 0 to 2pi} =>  2pi * H(x)

Integrate [ 27/4  sin(x) * (2pi) ] {p from 0 to pi}   =>
[ 27/4  * -cos(x) * (2pi) ]_x=0^\pi  => ...

-27 pi / 2  - - ( 27pi / 2 ) =  2 * (27/pi) =  27pi

Commented:
@Mysidia:

I'm having a hard time following the solution you posted. I'm not familiar with the notation you're using.
Commented:
What i'm saying is you only need a straightforward integration, a multiplication by 2pi, and another very straightforward integration.

It's not as if there are different orders to execute these steps in,  start evaluating the inner integral first,
then the middle one, then the outer integral.

a => b => c  is just a mixed up way of saying   a leads to b, and then a=c.

E(q)|_q=q0^q1   is just a shorthand for ' ( E(q1) - E(q0) )'.

Commented:
@Mysidia:

-27 pi / 2  - - ( 27pi / 2 ) =  2 * (27/pi) =  27pi

Commented:
Ok, last lines should in fact read

[ 27/4  * -cos(x) * (2pi) ] {from 0 to pi} =
(27/2 * -cos(pi) * pi)  - (27/2 * -cos(0) * pi)  =  2 * (27pi/2)
Commented:
I don't understand the complexity of the solutions posted so far.

Taking JR2003's point about concentric shells, each of area 4*pi*r² (using r rather than p for the radius), each of thickness dr (delta r) and of density (3-r) gives a mass of 4*pi*r²*(3-r)*dr. This needs to be integrated from r=0 (the middle) to r=3 (the outer limit).

The integration is quite simple and gives 4*pi*r^3 - pi*r^4
With the lower limit of zero this is zero, the upper limit is 4*pi*3^3 - pi*3^4
Factor out pi and 3 cubed (27) gives 27*pi*(4-3) or simply 27pi.

There is no need for triple integrals nor trigonometrical functions.
Commented:
Maybe Brian has a homework constraint that he has to use cylindrical or spherical coordinates.
Commented:
Brain actually wrote :-

Clearly the best option is a triple integral in spherical coordinates.

Now I've no idea where he got this from - it might even be his own - but I'll agrue the daylights out of ANY maths teacher about things like this. The BEST option is the SIMPLEST option. Mathematics is not about showing how clever we are, but about using tools to solve practical problems. There are relatively few problems which can be solved in spherical coordinates and those in 3-d Cartesian are also messy.

Commented:
agreed

Commented:
>>Maybe Brian has a homework constraint that he has to use cylindrical or spherical coordinates.

Well this problem was taken out of the section titled "Triple Integrals in Cylindrical and Spherical Coordinates", so I think it is implied that the problem _should_ be done in one of those two coordinate systems. I should have probably been clearer with this from the start, my apologies. My teacher provided his solution using a triple integral, however, using p - 3 vs 3 - p, that is really where my confusion is coming from.

-Brian

Commented:
However JR2003, NovaDenizen, and BigRat, you've all posted a method that leads to the correct solution.

I'm confused however by the fact that:

SA = 4pi*r^2
D = 3 - p

SA has units of (distance^2)
D has units of (mass/distance^3)

SA * D = MASS/DISTANCE .... not just mass as we are looking for. Now its clear that it works...i'm just confused how that part works.

-Brian

Commented:
The integral is of SA * dr, and dr has units (distance).

Commented:
I mean to say that the integral is of SA * D * dr, and the units there work out to MASS.  SA*dr is the infinitesimal volume of 'uniform' density that is being integrated.

Commented:
>>I mean to say that the integral is of SA * D * dr, and the units there work out to MASS.  SA*dr is the infinitesimal volume of 'uniform' density that is being integrated.

Makes sense, I forgot dr had units of distance.

Commented:
>>the problem _should_ be done in one of those two coordinate systems

Hmmmm. The problem is completely symetric about the origin. It is the worst example for triple integration I can think of. Presumably one considers a volume element at (r,theta,phi) with density 3-r and integrate all these over the spheroid.

I'll think out a better example for triple integration and post it as a question.

Commented:
>>Hmmmm. The problem is completely symetric about the origin. It is the worst example for triple integration I can think of.

How so, in spherical coordinates your phi would go from 0->pi, thetha 0->2pi, and your radius from 0->3, to me it seems pretty basic.

brian

Commented:
>>How so,

The density is proportional to the radius, ie: to r. The other two coordinates are not involved with the density only with the space, therefore the two integrations of theta and phi actually produce a shell of area 4*pi*r². As as question to show  triple integration using spherical coordinates it's weak to say the least.
Commented:
Just to prove a point, I'll partially solve it in spherical coordinates.  I'll do it with the p on the outermost integral.

x goes from 0 to pi, y goes from 0 to 2pi, p goes from 0 to 3
(3-p)p^2 (sin(x) dy dx) dp

While we integrate on x and y, the terms with p and dp are just multiplying constants so we can factor them out and ignore them for now.

sin(x) dy dx

Integrate on x
-cos(x) dy   from 0 to pi
(1 - -1) dy
2dy

Integrate this on y
2y from 0 to 2pi
4pi

Merging this back with the original integral, we get:
(3-p)4pi p^2 dp over p from 0 to 3
Which is exactly the simple integral I and others suggested.

Commented:
It's true that you don't need triple integration, if you happen to have the formula for the proper area available, the
extra two integrations just boil down to multiplication by the area.

However, I don't see anything erroneous with author's choice to set it up as a triple integral, from what I saw,
he did it correctly -- Whether it was a good choice or not is subjective.

If there was a technical error, it was almost certainly in the evaluation.

prog
Commented:
I vote for the autor's triple integral as a good and proper approach.

Commented:
>>I vote for the autor's triple integral as a good and proper approach.

I'll send you a sledge hammer for Christmas, so that you can crack your wallnuts.

Commented:
Ok, i've figured out this problem, will close soon.

Commented:
Thank you guys for your contributions to this problem.

Brian
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