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

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

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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Ask the ExpertsAnd 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?

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

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

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) )'.

[ 27/4 * -cos(x) * (2pi) ] {from 0 to pi} =

(27/2 * -cos(pi) * pi) - (27/2 * -cos(0) * pi) = 2 * (27pi/2)

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.

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.

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.

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.

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.

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.

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