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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?

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    $\begingroup$ I think its $r'$ just to destinguish from the bounds of the integral. $\endgroup$ – MrYouMath Aug 3 '16 at 19:21
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    $\begingroup$ 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 '16 at 19:28
  • $\begingroup$ @Dr.MV Thanks! I posted a follow-up question for this in my post. $\endgroup$ – JDoeDoe Aug 4 '16 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. $\endgroup$ – Christian Blatter Aug 4 '16 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 '16 at 11:52
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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.

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  • $\begingroup$ Thanks! I posted a follow-up question based on this in my post. $\endgroup$ – JDoeDoe Aug 4 '16 at 7:42

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