# Change in areal element

I am reading Griffith's Introduction to Electrodynamics., On example 1.7 while calculating surface integral of $$x = 2$$ for a cube of side 2., the book states $$da = dy \cdot dz$$ I don't get this, what I thought of is, $$a=y \cdot z \implies da = y \cdot dz + z \cdot dy$$ . Can anyone explain me what's wrong with this?

On using product rule, you get the small change in area (which is $$da$$) for a small change in $$y$$ (which is $$dy$$) and a small change in $$z$$ (which is $$dz$$). We ignore the $$dx\cdot dy$$ term because it is a second order differential which is negligible in comparison to other first order differentials. The image below will help you visualise this. The yellow area in the figure is the $$dx\cdot dy$$ term.

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But here in the question, our motive is to find an elemental area which can be used to represent the whole surface. We are not changing the original area in any way. We are just trying to represent a tiny elemental area so that we can apply the concepts of calculus. By elemental area, I mean something like this :-

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So here, when $$\Delta x \rightarrow 0$$ and $$\Delta y \rightarrow 0$$, we approach an infinitesimally small elemental area where we can consider the given vector function ($$\mathbf{v}=2xz \mathbf{\hat x}+(x+2)\mathbf{\hat y}+y(z^2-3)\mathbf{\hat z}$$) to be a constant vector.

You’ve computed the increase in the area of a rectangle of width $$y$$ and height $$z$$ when you increase the width by $$dy$$ and the height by $$dz$$. This is the growth on two sides. It is not the area of an infinitesimal rectangle of width $$dy$$ and height $$dz$$, which is obviously $$dy\,dz$$.