What is the relationship between dℓ and dR in the sign convention for voltage?

[Calpalned]

In the derivation for equation 24-2 in the picture below., the line integral was chosen along a path anti-parallel to the field lines. As a result, cos(180) = -1, which made the integral for the voltage positive.

However, in examples 24-2 and 24-3, the line integral is also taken as anti-parallel to the field lines, but the voltage integral remains negative. This seems contradictory.

 

[Simon Bridge]

Try doing the line integral yourself to check.
Where is the contradiction? The resulting voltage comes out positive.

 

[CrazyNinja]

The sign of voltage is more like a tool to decide direction of field. It's the magnitude we are more concerned with. So in either case just find the magnitude and the field is from +ve to -ve. Or you can take it as potential decreases on direction of field.
(I just said this to make stuff easy. The line integral will give you the right result always. Do as Simon said.)

 

[vela]

In Example 24-2, the integrand is $\vec{E}\cdot d\vec{l} = \|\vec{E}\| \|d\vec{l}\| \cos 180^\circ = -\|\vec{E}\| \|d\vec{l}\|$. The negative sign cancels with the negative sign in front of the integral, so you have
$$V = \int_a^b \|\vec{E}\| \|d\vec{l}\| > 0.$$ Note that $a$ and $R_a$ don't mean exactly the same thing; this is why the limits on the integral change as the integral is written in terms of $d\vec{l}$ and then in terms of $dR$. Now considering the fact that $R_a > R_b$, how is $\| d\vec{l} \|$ related to $dR$?