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vela
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In the integral
$$\int_1^0 dx,$$ what's the sign of dx?
$$\int_1^0 dx,$$ what's the sign of dx?
vela said:In the integral
$$\int_1^0 dx,$$ what's the sign of dx?
vela said:The integral is essentially a sum. By claiming dx>0 in that integral, you're saying by adding up a bunch of positive quantities, you can get a negative answer. You can't simply ignore the limits and assume dx is always positive.
vela said:It's because in those solutions, the dot product is found using ##\vec{F}\cdot d\vec{r} = F_x\,dx +F_y\,dy##. You don't discard the signs of the individual components when you use this formula. In your approach to the electric potential problem, however, you're using ##\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\|\cos \theta##. The negative sign arises because of the absolute value. The norm of the vector ##d\vec{r}## is given by ##\|d\vec{r}\| = \lvert dr\rvert = -dr## because ##dr<0##.
Say we want to calculate
$$\int_A^B \vec{F}\cdot d\vec{r}$$ where ##\vec{F} = \hat{i}## and points A and B lie, respectively, at (1,0) and (0,0).
We can parameterize the route ##\vec{r}(\lambda) = x(\lambda)\,\hat{i}+0\,\hat{j}## using ##x(\lambda) = 1-\lambda##, where the parameter ##\lambda## runs from 0 to 1. Then we have ##d\vec{r} = -d\lambda\,\hat{i}##. Here ##d\lambda>0## because the integral goes from ##\lambda=0## to ##\lambda=1##. This makes sense because ##d\vec{r}## should point in the -x direction when we move from x=1 to x=0.
If we calculate the dot product in terms of components, we get ##\vec{F}\cdot d\vec{r} = (-d\lambda)\,(\hat{i}\cdot \hat{i}) = -d\lambda##. The integral then becomes
$$\int_{\lambda=0}^{\lambda=1} -d\lambda = -1.$$ So far so good. If we use the other method to calculate the dot product, we have
$$\vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\| \cos \theta = 1\times\lvert d\lambda\rvert\times\cos 180^\circ = -\lvert d\lambda\rvert = -d\lambda,$$ so we'll end up with the same answer, as we should.
Now suppose instead we use the parameterization ##x(\lambda) = \lambda## where ##\lambda## goes from 1 to 0. This time, we'll get ##d\vec{r} = d\lambda\,\hat{i}##. If we calculate the dot product in terms of components, we get ##\vec{F}\cdot d\vec{r} = d\lambda\,(\hat{i}\cdot \hat{i}) = d\lambda##. The integral then becomes
$$\int_{\lambda=1}^{\lambda=0} d\lambda = -1.$$ There's no need to take into account the signs of the components. It all works out automatically.
If we use the other method to calculate the dot product, we have
$$\int_{\lambda=1}^{\lambda=0} \vec{F}\cdot d\vec{r} =
\int_{\lambda=1}^{\lambda=0} \|\vec{F}\| \|d\vec{r}\| \cos 180^\circ =
\int_{\lambda=1}^{\lambda=0} -\lvert d\lambda\rvert.$$ From this, you can clearly see we must have ##\lvert d\lambda\rvert = -d\lambda## otherwise we get the wrong sign on the final result. But this exactly what we should expect because ##d\lambda<0## with this parameterization of the path.