# Surface Integral Equations

I am studying for a Calculus exam, and one of the topics I should know about is surface integrals. Now, I am using Stewart 6e, and in there I have found several equations for computing surface integrals. These are the equations I found:

$$\iint_{S}{f(x, y, z)\:dS}=\iint_{D}{f(\vec{r}(u,v))*\left|\vec{r}_{u}\times\vec{r}_{v}\right|\:dA} \\ \iint_{S}{f(x, y, z)\:dS}=\iint_{D}{f(x, y, g(x, y))*\sqrt{\left(\frac{dg}{dx}\right)^{2}+\left(\frac{dg}{dy}\right)^{2}+1}\:dA} \\ \iint_{S}{\vec{F}\:d\vec{S}}=\iint_{D}{\vec{F}*\left(\vec{r}_{u}\times\vec{r}_{v}\right)\:dA} \\ \iint_{S}{\vec{F}\:d\vec{S}}=\iint_{D}{\left(-P\left(\frac{dg}{dx}\right)-Q\left(\frac{dg}{dy}\right)+R\right)\:dA}$$

So, my question is: How do I know which one to use for a given exercise? Obviously the bottom two should be used if I'm dealing with a vector field, and the top two if I'm dealing with just an equation. Also, I noticed that the second and last one both require a seperate function g(x,y). But still, it isn't quite clear to me when to use which one. Please help me out!

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It all comes down to how the surface you integrate over is most easily described. Number one and three are for surfaces described by a parametrization, as in you have one vector function $\vec{r}(u, v)$ giving you the full three coordinates of each point. Number two and four is for when the surface is the graph of the function $g(x, y)$, where you supply and $x$ and $y$ coordinate, and then $g$ tells you what the corresponding $z$-coordinate is. – Arthur Jun 24 '12 at 11:25
@Legion Several users added LaTeX to you post based on the picture you originally posted. Please, check whether it is of and, if needed, make further edits to the LaTeX code. – Martin Sleziak Jun 24 '12 at 11:28
@Arthur: Thank you very much for your quick response, but I have a question about your answer. What is the definition of a "graph"? What are the criteria for something to be a "graph"? I have asked my teacher this, but his answer was very confusing and thus not very helpful (although it was well-intended). @Martin Sleziak: Yes I noticed, and I'd like to thank them for doing this, since I don't know how to use LaTeX yet. It is all still correct as far as I can tell. – Legion Jun 24 '12 at 11:31
In this case, a graph is a visualisation of a function taking two variables, and producing a single real number output. To give some examples, the graph of $g(x, y) = 1$ is a horizontal plane intersecting the $z$-axis at $z=1$. The graph of $h(x, y) = \sqrt{x^2 + y^2}\:$ is a cone with its tip in the origin, and expanding upwards. – Arthur Jun 24 '12 at 11:38
Ok I think I see now, the difference is then whether or not one is directly integrating over an arbitrary surface in space or over the surface of another graph. And then there's the division between whether or not you're dealing with a "normal" function or a vector field. Is that correct? – Legion Jun 24 '12 at 11:47

When there is $\hat n$ involved, use $$\iint_s \vec F(x,y,g(x,y))\cdot \frac{\nabla u(x, y)}{|\nabla u(x, y)|} \sqrt{ 1 + \left ( \partial g \over \partial x\right )^2 + \left ( \partial g \over \partial y\right )^2} dxdy$$ $u$ being surface. Or, $$\iint_{S} \vec F(x, y, z). \hat n dS=\iint_{D}{ \vec F(\vec{r}(u,v))*\left(\vec{r}_{u}\times\vec{r}_{v}\right)\:dudv}$$
If there is not $\hat n$
$$\iint_s \vec F(x,y,g(x,y)) \sqrt{ 1 + \left ( \partial g \over \partial x\right )^2 + \left ( \partial g \over \partial y\right )^2} dxdy$$ Or, $$\iint_{S} \vec F(x, y, z) dS=\iint_{D}{ \vec F(\vec{r}(u,v))\left|\vec{r}_{u}\times\vec{r}_{v}\right|\:dudv}$$ Both are equivalent. check this out surface integral of vector along the curved surface of cylinder and parametrization of surface element in surface integrals.