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

## Homework Statement

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.

## The Attempt at a Solution

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Mark44
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This way? No sure is that correct .

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Mark44
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LCKurtz
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## Homework Statement

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.

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.

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?).

(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.

LCKurtz
Homework Helper
Gold Member
I thought you said you were puzzled about 1, not 2.

(1/r)(dr/dt)= 0 , (dr/dt)= 0
∫ dr = ∫ 0 dt
The answer should be r = constant.
Isn't ?

Yes, that's right.

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.

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

I got it. Thanks much