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I'm doing some surface integrals and I'm confused as to which formula I should use.

Let's say you have a parametric representation of a surface S.

$\overrightarrow{r}(u,v) = x(u,v)\overrightarrow{i} + y(u,v)\overrightarrow{j} + z(u,v)\overrightarrow{k}$

That is, from what I can gather, taking some 2-dimensional representation of a surface and extending it outward into a 3-dimensional space. And the equation above gives us the vector $\overrightarrow{r}$ that gives the x, y, and z coordinates of the point on the 3D surface.

After computing two tangent vectors, $r_u$ and $r_v$, you can take the cross product to get the area of a parallelogram spanned out by those two vectors.

$\iint_D ||r_u \times r_v||~dA$

This is supposed to give us the surface area of that 2D surface that has been extended out into the third dimension. Right?

But where I get confused is when my text introduces a completely different equation for the surface area for a surface z = g(x,y).

$\iint_S f(x,y,z)dS = \iint_D f(x,y,g(x,y))\sqrt{(\frac{\partial g}{\partial x})^2 + (\frac{\partial g}{\partial y})^2 + 1} dA$

When are you supposed to use each of these equations? Is the first when you have a two dimensional surface that is extended out into the third dimension, and the second one is when you have just a straight up 3D surface?

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2 Answers

My advice is to minimize the number of vector integral formulas that you keep in memory. All you really need for surfaces are the two definitions $$\iint_S f\,dS = \iint_D f(\vec r)\|\vec r_u\times \vec r_v\|\,dA, \qquad \iint_S \vec F\cdot \overrightarrow{dS} = \iint_D F(\vec r)\cdot (\vec r_u\times \vec r_v)\,dA$$ plus the computational shortcut: if $r=x\vec i+y\vec j+g(x,y)\vec k$, then $\vec r_x\times \vec r_v = -g_x \vec i-g_y \vec j+\vec k$ (notably, this is an upward normal). You should also know that the length/area/volume of anything is the integral of $1$ over that object.

The two formulas for surface integrals should be compared to the similar line integral formulas: this makes the set of four easier to remember. $$\int_C f\,ds = \int f(\vec r)\|\vec r_t\|\,dt, \qquad \int_C \vec F\cdot \overrightarrow{dr} = \int F(\vec r)\cdot \vec r_t\,dt $$

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I'm not sure I understand your concept of "extending a surface out into the third dimension". I would describe the parametric surface differently: A domain $D$ of the two-dimensional space of the parameters $u,v$ is mapped to a surface in three-dimensional space; that is, the parametric representation assigns to each point of $D$ a point in $\mathbb R^3$, and the set of all these points, i.e. the image of $D$ under this map, is a two-dimensional surface in three-dimensional space.

Your last equation actually gives a formula not for the area of the surface, but for the integral of a function over the surface. You'd get the area by choosing $f\equiv1$. Likewise, you could put a function $f(\vec r(u,v))$ into your first formula to get the integral of a function over the surface instead of the area.

To understand the relationship between the two formulas (with or without $f$), it's important to understand what $\mathrm dA$ and $\mathrm dS$ are: $\mathrm dA$ is the area element in parameter space, whereas $\mathrm dS$ is the area element of the surface in three-dimensional space. They're related by $\mathrm dS=\lVert\vec r_u\times\vec r_v\rVert\mathrm dA$. (You can get the right spacing for the norm bars by using \lVert and \rVert instead of double bars.)

What may be confusing about your second approach is that here $x$ and $y$ play a double role – they're two of the three coordinates in the three-dimensional space, but they're also serving as the two parameters. Written out explicitly, this would be

$$ \vec r(x,y)=\pmatrix{x\\y\\g(x,y)}\;, $$

$$ \vec r_x=\pmatrix{1\\0\\\partial g/\partial x}\;, $$

$$ \vec r_y=\pmatrix{0\\1\\\partial g/\partial y}\;, $$

$$ \vec r_x\times\vec r_y=\pmatrix{-\partial g/\partial x\\-\partial g/\partial y\\1}\;, $$

$$ \lVert\vec r_x\times\vec r_y\rVert=\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}\;. $$

So the second formula is a special case of the first one, in which the parameters happen to be the coordinates $x$ and $y$. This is often useful if the surface is given in terms of $z$ as a function of $x$ and $y$.

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