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?
Best Answer
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. $$\int_C f\,ds = \int f(\vec r)\|\vec r_t\|\,dt, \qquad \int_C \vec F\cdot \overrightarrow{dr} = \int F(\vec r)\cdot \vec r_t\,dt $$