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Consider the scalar surface integral:

$$\iint_{S}G(x, y, z) dS=\iint _R\left(G(x, y, f(x,y))\cdot\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\right)dxdy$$

where $S\subset z=f(x,y)$ and $R$ is the projection of $S$ onto the xy plane.

The above formula gives a method for evaluating the surface integral in terms of a standard double integral.

One thing that struck me as interesting, however, was the $\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}$ multiplicand. This seems eerily similar to the arc-length integrand, except one-dimension lower. This similarity is also thus manifested in the surface area of revolution formula.

Is there any deeper connection between why this appears in both?

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  • $\begingroup$ Am I correct in intuiting that $\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dx dy$ acts sort of as a "surface element" just as the lower-dimension version acts as an arclength differential? $\endgroup$ – 1110101001 Dec 21 '15 at 4:06
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Yep, your intuition is bang on. For an arc-length integral, we would compute $$ \int_\gamma \mathrm{d} s = \int_\gamma \sqrt{\mathrm{d}x ^2 + \mathrm{d}y^2 }. $$ In some sense, this arc length element $\mathrm{d}s$ comes from the Pythagorean theorem.

If $\gamma$ is given by the graph $y=f(x)$, we note that $\mathrm{d} y = \mathrm{d} f(x) = f'(x) \mathrm{d}x$. Our integral becomes $$\int_\gamma \mathrm{d}s = \int_a^b \sqrt{\mathrm{d}x^2+ ( f'(x) \mathrm{d}x)^2} = \int_a^b \sqrt{1+ (f'(x))^2} ~\mathrm{d}x.$$

Moving to the surface area integral of the graph $z=f(x,y)$, we note that $\mathrm{d}z = \mathrm{d} f(x,y) = f_x \mathrm{d}x + f_y \mathrm{d} y$. Using that the area of a parallelogram spanned by $\vec{v}$ and $\vec{w}$ is $\| \vec{v} \times \vec{w} \|$, we have that an "infinitesimal parallelogram" on the surface is spanned by $(1,0, f_x)$ and $(0,1,f_y)$.

\begin{align*} \iint_M \mathrm{d}S &= \iint_M \|(1,0,f_x) \times (0,1,f_y) \| \mathrm{d}x \mathrm{d}y \\ &= \iint_M \| (-f_x, -f_y, 1) \|\mathrm{d}x \mathrm{d}y \\ &= \iint_M \sqrt{(f_x)^2 + (f_y)^2 + 1 } ~\mathrm{d}x \mathrm{d}y. \end{align*} Notice that we again computed the length of something using the Pythagorean theorem; this is no accident! This story actually goes even deeper, but would require a knowledge of differential forms / exterior derivatives / wedge products. In some sense, those subjects are about a connection between the gradient, curl, and divergence operators and a desire to only ever use $\mathrm{d}$, instead of $\nabla f$, $\nabla \times \vec{F}$, and $\nabla \cdot \vec{F}$. See this summary for the curious / brave. (Edit: additionally, this writeup gives a "reasonable" introduction to differential forms / tensor calculus written at the undergraduate level.)

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  • $\begingroup$ Interesting — could you please elaborate more on the deeper connection using differential forms? (I probably won't understand it yet but I'd like to have something to come back to once I read up on them) $\endgroup$ – 1110101001 Mar 30 '17 at 5:05
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    $\begingroup$ See the link in the last paragraph for an intro to differential forms; also math.purdue.edu/~dvb/preprints/diffforms.pdf is a nice exposition of differential forms written for the undergraduate level, around page 24 has the answer to your question. $\endgroup$ – erfink Mar 30 '17 at 5:22

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