Scalar surface integral with prime symbol, why? 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?

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.
