# How to integral ? How can they ignore that variable?

1. Jul 25, 2013

### Outrageous

1. The problem statement, all variables and given/known data
I don't understand case 1. In case 2 that is normal integration , isn't?
Then why case 1 can simply ignore the r ? I don't think r is constant.

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Jul 25, 2013
2. Jul 25, 2013

### Staff: Mentor

Last edited: Jul 25, 2013
3. Jul 25, 2013

### Outrageous

This way? No sure is that correct .

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4. Jul 25, 2013

### Staff: Mentor

5. Jul 25, 2013

### Outrageous

Sorry, I don't really know how to use it.

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6. Jul 25, 2013

### LCKurtz

Are you talking about the equation$$\frac m r \frac{d(r^2\dot \theta)}{dt}=0$$giving $r^2\dot \theta=l$? If so, they didn't ignore the $\frac m r$. They multiplied both sides by its reciprocal before integrating.

7. Jul 25, 2013

### Boorglar

I'm not quite sure what you mean. Nobody said r was constant. But the fact that $\frac{d}{dt}(r^2\theta') = 0$ does imply that $r^2\theta' = C$ is constant (why?).

8. Jul 25, 2013

### Outrageous

(1/r)(dr/dt)= 0 , (dr/dt)= 0
∫ dr = ∫ 0 dt
The answer should be r = constant.
Isn't ?
Can the answer be ∫(1/r)dr= ∫dt .
Then ln(r/ro)= t. Integral from ro to r , and integral from 0 to t. ro is original value r.

9. Jul 25, 2013

### LCKurtz

I thought you said you were puzzled about 1, not 2.

Yes, that's right.

$\int \frac 1 r\, dr = \int 0\, dt$, not $\int 1\, dt$.

10. Jul 26, 2013

### Outrageous

I got it. Thanks much
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