How Do I Integrate This Sphere-Related Equation Correctly?

[jderulo]

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

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

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

I've had a go but unsure what to do with the $r^2$

I thought this might work but it gives a negative value for $q_m$

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

Integrated to $q_m ln \frac{r_2}{r_1^2} = 4 \pi D (c_1 - c_2)$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 $$4\pi r^2$$ so that $q_m= -DA\frac{dc}{dr}$ (NOT $\frac{dc}{dx}$) becomes $q_m= -D(4\pi r^2)\frac{dc}{dr}$ which can be separated as
$$\frac{q_m}{r^2}dr= -4D\pi dc$$
and integrating,
$$q_m\int_{r_1}^{r_2} \frac{1}{r^2}dr= -4D\pi\int_{c_1}^{c_2} dc$$

Integrate that.

 

[jderulo]

This?

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

 

[HallsofIvy]

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

 

[jderulo]

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

How do I fit in $r_1$ and $r_2$

 

[HallsofIvy]

If $\int f(x)dx= F(x)+ C$ then
$$\int_{x_1}^{x_2} f(x)dx= F(x_2)- F(x_1)$$.

 

[jderulo]

If $\int f(x)dx= F(x)+ C$ then
$$\int_{x_1}^{x_2} f(x)dx= F(x_2)- F(x_1)$$.

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