Computing Dot Product: (\nabla\times \mathbf{v})\cdot d\mathbf{a}

[Saladsamurai]

I cannot seem to figure out how to compute this dot product?!

If (\nabla\times \mathbf{v})=(4z^2-2x)\hat{i}+2z\hat{k} and d\mathbf{a}=dydz\hat{i}

Then shouldn't the DOT PODUCT be:

(\nabla\times \mathbf{v})\cdot d\mathbf{a}=(4z^2-2x)\hat{i}*dydz\hat{i}=(4z^2-2x)dydz ?

But the book says its just 4z^2dydz

What am I doing wrong here??

Here is the original question:

 

[Saladsamurai]

Wait... I think I see what he did. Since the square lies in the z-y plane only, when the integral is carried out, anything with an x in it will vanish anyway. So he just did not bother.

I really hate the way this guy does his math.

 

[D H]

He did bother. He said "Since x=0 for this surface" right before he wrote out the integral.

Look at it this way: If authors had to spell out every step along the way they wouldn't be able to get past introductory algebra, let alone discuss Stokes' theorem. Authors have to omit obvious intermediate steps. Some authors see things as obvious that the novice doesn't. This, IMO, is not one of those cases. If the author had failed to say "Since x=0 ..." you would have had a better case.

 

[Saladsamurai]

Point taken. But mind you, my malcontent is not a result of his omitting obvious steps. He could have just computed the dot product, ran the integrals, and left to to the reader to figure out why all of the x terms dropped out.

Different strokes for different folks. I've just never seen it done this way before