Integration - slightly confused

[jderulo]

Hi I'm trying to integrate the following [itex]q_m = -D A \frac{dc}{dx}[/itex]
where [itex]A = 4 \pi r^2[/itex] Yes, a sphere.My supplied literature simplifies to [itex]q_m = -D 2 \pi r L \frac{dc}{dr}[/itex] when [itex]A = 2 \pi r L [/itex]

Integrating to [itex]\int_{r1}^{r2} q_m \frac{dr}{r} = - \int_{c1}^{c2} 2 \pi L D dc[/itex]

Integrated to [itex]q_m ln \frac{r_2}{r_1} = 2 \pi L D (c_1 - c_2) [/itex]

I've had a go but unsure what to do with the [itex]r^2[/itex]

I thought this might work but it gives a negative value for [itex]q_m[/itex]

[itex]\int_{r1}^{r2} q_m \frac{dr}{r^2} = - \int_{c1}^{c2} 4 \pi D dc[/itex]

Integrated to [itex]q_m ln \frac{r_2}{r_1^2} = 4 \pi D (c_1 - c_2) [/itex]Any ideas? Thanks.

 

[Nick O]

I don't follow... how does x relate to r?

 

[jderulo]

I don't follow... how does x relate to r?

I've edited it whilst you were reading. Do you follow now? They replace x with r.

 

[Nick O]

Not really. Can you post the problem statement? It isn't even clear that r is a variable, or that there is correlation between x and r.

 

[jderulo]

I've attached the literature. Basically the formula is for a cylinder, I need to convert it for a sphere.

 

[HallsofIvy]

What you post in your last response integrates over a cylinder of radius r, length L. But you are asking about a sphere. Are you asking how to change from a cylinder to a sphere? If so then, yes, the surface area of a sphere of radius r is [tex]4\pi r^2[/tex] so that [itex]q_m= -DA\frac{dc}{dr}[/itex] (NOT [itex]\frac{dc}{dx}[/itex]) becomes [itex]q_m= -D(4\pi r^2)\frac{dc}{dr}[/itex] which can be separated as
[tex]\frac{q_m}{r^2}dr= -4D\pi dc[/tex]
and integrating,
[tex]q_m\int_{r_1}^{r_2} \frac{1}{r^2}dr= -4D\pi\int_{c_1}^{c_2} dc[/tex]

Integrate that.

 

[jderulo]

This?

[itex]q_m ln (\frac{1}{r_1^2}r_2) = 4 \pi D (c_1 - c_2) [/itex]

 

[HallsofIvy]

Then you are not "slightly confused" about integration- you seem to be saying you do not know how to integrate at all. No [itex]\int dr/r^2[/itex] is not "ln(r^2)". Using the very basic rule [itex]\int r^n dr= r^{n+1}/(n+ 1)[/itex], [itex]\int dr/r^2= \int r^{-2}dr= -r^{-1}+ C[/itex].

 

[jderulo]

Then you are not "slightly confused" about integration- you seem to be saying you do not know how to integrate at all. No [itex]\int dr/r^2[/itex] is not "ln(r^2)". Using the very basic rule [itex]\int r^n dr= r^{n+1}/(n+ 1)[/itex], [itex]\int dr/r^2= \int r^{-2}dr= -r^{-1}+ C[/itex].

How do I fit in [itex] r_1 [/itex] and [itex] r_2 [/itex]

 

[HallsofIvy]

If [itex]\int f(x)dx= F(x)+ C[/itex] then
[tex]\int_{x_1}^{x_2} f(x)dx= F(x_2)- F(x_1)[/tex].

 

[jderulo]

If [itex]\int f(x)dx= F(x)+ C[/itex] then
[tex]\int_{x_1}^{x_2} f(x)dx= F(x_2)- F(x_1)[/tex].

Thanks, I thought as much but seeing as I made a glaring error previously I thought I should double check.