Surface Integrals: Online Guide to Solving 3D Problems

[theneedtoknow]

Can someone either explain here, or link me to an online document, on how to do surface integrals over surfaces in 3d (not simple ones like planes with x, y, or z held constant). I learned this back in Calculus 2 five years ago, and now I need to do it for my Electrodynamics course and I can't really remember how to do it?

For example, what if I have a function that I want to integrate over a triangle with vertices (1, 0, 0), (0, 2, 0), (0, 0, 3)?
I think it had something to do with with finding a unit vector normal to the surface (which I can do), but no idea where to go from there.

 

[HallsofIvy]

Can someone either explain here, or link me to an online document, on how to do surface integrals over surfaces in 3d (not simple ones like planes with x, y, or z held constant). I learned this back in Calculus 2 five years ago, and now I need to do it for my Electrodynamics course and I can't really remember how to do it?

For example, what if I have a function that I want to integrate over a triangle with vertices (1, 0, 0), (0, 2, 0), (0, 0, 3)?
I think it had something to do with with finding a unit vector normal to the surface (which I can do), but no idea where to go from there.

Here is the standard thing to do.
The vector from (1, 0, 0) to (0, 2, 0) is <0- s, 2- 0, 0-0>= <-1, 2, 0> and that is a vector in the plane. The vector from (1, 0, 0) to (0, 0, 3) is <0- 1, 0- 0, 3- 0>= <-1, 0, 3> and that is also a vector in the plane. Any point in the plane can then be written as (1, 0, 0)+ t(-1, 2, 0)+ s(-1, 0, 3): x= 1- t- s, y= 2t, z= 3s. For example, if t= s= 0, that is (1, 0, 0), if t= 1, s= 0, that is (0, 2, 0), and if t= 0, s= 1, that is (0, 0, 3). The position vector of any point in the plane can be written as [itex]\vec{r}= <1- t- s, 2t, 3s>[/itex]. Differentiating with respect to t and s we get [itex]\vec{r}_t= <-1, 2, 0>[/itex] and [itex]\vec{r}_s= <-1, 0, 3>[/itex] which are, of course, just the vectors between points we had before. The "fundamental vector product" for a surface is [/itex]\vec{r}_t\times \vec{r}_s[/itex], the cross product of those derivatives. The "differential of surface area" is just [itex]|\vex{r}_t\times\vec{r}_s|dsdt[/itex].

For this problem, where the surface is a plane, that is just the cross product of the two vectors, <-1, 2, 0> and < -1, 0, 3>, from (1, 0, 0) to the other two points: <6, 3, 2>. Its length is [itex]\sqrt{36+ 9+ 4}= \sqrt{49}= 7[/itex]. (for a plane that is always a constant, not necessarily an integer!) The differential of surface area is 7dsdt.

To integrate, say, f(x,y,z) over a region on that plane, replace x by 1- t- s, y by 2t, and z by 3s and integrate [itex]7\int f(1- t- s, 2t, 3s) ds dt[/itex].

 

[theneedtoknow]

Thank you! How do I go about figuring out the limits of integration of s and t?

 

[theneedtoknow]

Also , if my function f is a vector function, how do i set up the integral?
I need to dot the vector function with the area differential da, so I can't just multiply it by Cdsdt

 

[LCKurtz]

Please do not waste our time by posting the same question in two different categories.

 

[theneedtoknow]

I apologize for the inconvenience, it's just that when i posted this here, I had no idea how the process even worked, so it seemed valid to post in here to get a general procedure for this. After I tried applying the process the person in this topic described to a specific problem, and ran into difficulty, it seemed inappropriate to post it on this forum since it says "no homework questions", so I started a new topic in the homework section.

 

[HallsofIvy]

Plot the given points in "st-space". From the way we got them, s= t= 0 is the point (1, 0, 0); s= 0, t= 1 is the point (0, 2, 0); and s= 1, t= 0 is the point (0, 0, 3). The triangle having vertices (1, 0, 0), (0, 2, 0), and (0, 0, 3) corresponds to the triangle in "st-space" with vertices (0, 0), (1, 0), and (0, 1). The boundaries are, of course, the lines s= 0, t= 0, and s+ t= 1.

To integrate over that region you could, for example, take s going from 0 to 1. Then, for each s, t goes from 0 to t= 1- s. That would give the integral
[tex]\int_{s=0}^1\int_{t=0}^{1- s} f(s,t)(7dtds)[/tex].

Or you could take t going from 0 to 1 and, for each t, s from 0 to 1- t. That would give
[tex]\int_{t=0}^1\int_{s=0}^{1- t} f(s,t)(7dsdt)[/tex]

 

[theneedtoknow]

Thank you for the help:)

 

[HallsofIvy]

To integrate a vector function, use the "vector" differential of surface area, [tex]d\vec{S}[/tex] which is the [itex]\vec{r_s}\times\vec{r_t} dsdt[/itex] rather than its length as I used before. Take the dot product of the vector function with that.

To integrate a vector function, [itex]f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex], over the plane with vertices at (1, 0, 0), (0, 2, 0), and (0, 0, 3) as in your first post, you take the dot product of that vector with the vector <6, 3, 2>, whose length we used before. The integral would be
[tex]\int_{s=0}^1\int_{t= 0}^{1- s} (f(1-s-t,2t,3s)+ 3g(1-s-t,2t,3s)+ 2h(1-s-t,2t,3s) )dtds[/tex]