@WhyWhatWhereWhenHow I don't cry, 'cause I haven't done it yet :P

@Charlie Now all are multiples of 10 :D

@MadameAkira lol, how can it be late. It was barely 7 pm when you said that... Good Night at 7 pm, doesn't suit a physics student :P

@WhyWhatWhereWhenHow the oscillations in a transverse wave are perpendicular (transverse) to the direction of propagation, whereas the oscillations in a longitudinal wave re along the direction of propagation of the wave.

@ACuriousMind Can we use integration by parts to integrate the dot product of two vectors. Something like this: $$\int \mathbf a \cdot \mathrm d \mathbf b=\mathbf a\cdot\mathbf b-\int \mathbf b\cdot \mathrm d \mathbf a$$

@ACuriousMind oops! I did just the same thing, sorry.

Answer if convenient...

@ACuriousMind yeah, I am integrating the dot product of an infinitesimal vector and a finite vector. But apart from the sloppy terminology, is the math right?

@ACuriousMind $\mathbf b$ is a vector such that $\mathbf a$ is a function of $\mathbf b$.

@ACuriousMind Oh yeah, in the original doubt I wanted to integrate the normal vector of a surface dotted another vector over that surface.

@ACuriousMind Hmmm... I know Stokes' theorem, though I am not at all familiar with the topological terminology you just used :)

@ACuriousMind what is /\ ?

@ACuriousMind hmm... Looking it up...

@ACuriousMind so if I want to integrate a force $\mathbf a$ (to find the potential difference) along a straight line from $0$ to $\mathbf b$, such that $\mathbf a=f(\mathbf b)$, where $f:\mathbb R^3\mapsto \mathbb R^3$, how do I do that?

Would this formula work?

@ACuriousMind that's what I am probably asking. Is it legal to transform a line integral like I did.

@ACuriousMind we can assume $\mathbf a=f(\mathbf b)=\lambda \mathbf b$ for simplicity...

@ACuriousMind just for learning purposes, just this once...

@ACuriousMind :P Well, then...

@ACuriousMind alright, let me include limits...

$$\int_{0}^{\mathbf b_0}\mathbf a \cdot\mathrm d \mathbf b=\mathbf a\cdot\mathbf b\biggr|_{0}^{\mathbf b_0}-\int_{0}^{\mathbf b_0} \mathbf b\cdot \mathrm d \mathbf a$$

@Yuvraj no. This case is just an easy example chosen by me. I want the general way to integrate such a dot product.

@ACuriousMind could you please pinpoint the exact error?

@ACuriousMind Oops!My bad.1.$\mathbf b_0$ is a fixed vector in space.2.$f'(\mathbf b)\mathrm d \mathbf b$ or in other words, the infinitesimal change in $\mathbf a$ along the straight line3.That was my mistake. I didn't see that we needed to talk about limits as well, so I just thought to use $\mathrm d \mathbf b$ as an infinitesimal vector along the straight line from $0$ to $\mathbf b_0$. From now on I'll stay consistent with this notation. 4. I don't really understand what you're trying to say...

@ACuriousMind But isn't $g' \mathrm dx$ equivalent to $\mathrm d g$?

@ACuriousMind $\mathbf b_0$ defines the end of the line, whereas, $\mathbf b$ is our variable.

@ACuriousMind So, is there a problem in my application of chain rule, or is there a problem in my definition of various parameters.

@ACuriousMind yes, I don't mean it.

@ACuriousMind it's the vector along the line joining $0$ and $\mathbf b_0$, and it runs from $0$ to $\mathbf b_0$. Thus $\mathrm d \mathbf b$ is an infinitesimal change along the line joining $0$ and $\mathbf b_0$.

@ACuriousMind Oops! Then I suppose I have been doing some illegal stuff lately :P

@ACuriousMind yeah, It seems I am assuming some weird stuff without involving the necessary rigour.

@ACuriousMind alright, I agree. So let's just assume that $\mathrm d \mathbf b$ is a tangent vector to the line at each point, what next?

@ACuriousMind Like compute the dot product at each point and then do the normal "scalar" integration, right?

@ACuriousMind so, in the above example, since $\mathrm d\mathbf b=\mathrm d \mathbf r$, and assuming $\mathbf a =\lambda \mathbf b =\lambda \mathbf r$, we get:$$\int_0 ^{\mathbf b_0}\mathbf a\cdot \mathrm d\mathbf b=\int_0 ^{\mathbf b_0} \lambda \mathbf r\mathrm d\mathbf r=\int_0 ^{|\mathbf b_0|} \lambda r\mathrm dr=\frac{\lambda r^2}{2}\biggr|_0 ^{|\mathbf b_0|}=\frac{\lambda |\mathbf b_0|^2}{2}$$ Is this correct?

@ACuriousMind I think that I kinda know what a line integral is, but I was just messing around...