1
$\begingroup$

What is the difference between $\int_0^{2\pi} \int_{-r}^{r} r\, dr\, d\phi$ and $\int_0^{2\pi} \int_{-r}^{r} r'\, dr'\, d\phi?$ Why the prime symbol?

What is the geometric meaning of $r'$ (in the problem below)?

I wondering because of this problem: Evaluate the scalar surface integral $\iint_S dS$ where $S: x^2+y^2=r^2$.

Solution: Parameterize the circle: $x=r\cos\phi$ and $y=r\sin\phi$: $$\mathbf{r}(r,\phi)=r\cos\phi{\mathbf{\hat{x}}}+r\sin\phi{\mathbf{\hat{y}}}+0{\mathbf{\hat{x}}}$$ Where $r\in[-r,r]$ and $\phi\in[0,2\pi]$. $$ \frac{\partial\mathbf{r}(r,\phi)}{\partial r}=\cos\phi{\mathbf{\hat{x}}}+\sin\phi{\mathbf{\hat{y}}}+0{\mathbf{\hat{x}}} $$ $$ \frac{\partial\mathbf{r}(r,\phi)}{\partial \phi} =-r\sin\phi{\mathbf{\hat{x}}}+r\cos\phi{\mathbf{\hat{y}}}+0{\mathbf{\hat{x}}} $$ $$ \bigg \lvert \frac{\partial\mathbf{r}(r,\phi)}{\partial r} \times \frac{\partial\mathbf{r}(r,\phi)}{\partial \phi} \bigg \rvert = r $$ And $$ dS=\bigg \lvert \frac{\partial\mathbf{r}(r,\phi)}{\partial r} \times \frac{\partial\mathbf{r}(r,\phi)}{\partial \phi} \bigg \rvert \, dr \, d\phi = r\, dr\, d\phi $$ So $$ \iint_S dS = \int_0^{2\pi} \int_{-r}^{r} r\, dr\, d\phi \tag{1} $$ However, the integral in my book is denoted $$ \iint_S dS = \int_0^{2\pi} \int_{-r}^{r} r'\, dr'\, d\phi \tag{2} $$ What is the difference? The radius is $r$, but what is the geometric meaning of $r'$?

EDIT: Follow-up question:

From the answers $r'$ is just to distinguish the dummy/integration variable from the bounds. However, in the parameterization $r$ is the radius of the circle, but instead using $r'$ we have $$x=r'\cos\phi \quad \text{and} \quad y=r'\sin\phi$$ But then $r'^2=r^2$?

I don't grasp it, how to think about it?

$\endgroup$
5
  • 1
    $\begingroup$ I think its $r'$ just to destinguish from the bounds of the integral. $\endgroup$
    – MrYouMath
    Aug 3, 2016 at 19:21
  • 3
    $\begingroup$ The primed coordinates are dummy integration variables while the unprimed ones are fixed parameters not implicated in the integration process. $\endgroup$
    – Mark Viola
    Aug 3, 2016 at 19:28
  • $\begingroup$ @Dr.MV Thanks! I posted a follow-up question for this in my post. $\endgroup$
    – JDoeDoe
    Aug 4, 2016 at 7:41
  • $\begingroup$ Your $S: \ x^2+y^2=1$ is a curve in the $(x,y)$-plane (with zero area), or an infinite cylinder in $3$-space, hence has infinite area. $\endgroup$ Aug 4, 2016 at 8:33
  • $\begingroup$ Here $\phi$ goes from $0$ to $2\pi$, so they didn't (although they could have) write $d\phi'$. $\endgroup$
    – GEdgar
    Aug 4, 2016 at 11:52

1 Answer 1

1
$\begingroup$

It is thought that using $r$ for bound and unbound variables at the same time is bad notation. So $$ \int_{-r}^r r\;dr $$ should be avoided.

$\endgroup$
1
  • $\begingroup$ Thanks! I posted a follow-up question based on this in my post. $\endgroup$
    – JDoeDoe
    Aug 4, 2016 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.