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?

  • $\begingroup$ 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

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.)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.