I am a lot of problems with parametrizing surfaces

[flyingpig]

Homework Statement

Evaluate the surface integral

$$\iint_S xy \;dS$$

S is the boundary is the boundary of the region enclosed by the cylinder $$x^2 + z^2 =1$$ and the planes y = 0 and x + y = 2


2. The Solutions

[PLAIN]http://img844.imageshack.us/img844/8827/unledje.jpg

The Attempt at a Solution

I honestly don't understand a thing in the solutions, especially the parametrization. I just don't even know where to begin. I know the enclosed surface, but I just don't know the math to write it out.

I can compute the integral fine, but I can't parametrize the surface.

I drew a picture, but I can't upload it.

This is what I have so far

$$y \in [0,2-x]$$

$$x \in [-1,1]$$

And I don't have a good clue about the ranges for the z values...

 

[JThompson]

Based upon the intervals you wrote down, it looks like you are trying to find a single parametrization for the whole surface- don't try that. This will not work because the pieces of the surface, i.e. the cylinder and the two planes, have different normal vectors.

1) Do you see why they broke S up into 3 separate surfaces?
2) Do you know how to parametrize a cylinder (whose axis lies on a Cartesian coordinate axis) and a plane?

To parametrize a surface you need two parameters.
One of the planes is y=0, which means that x and z are independent variables. So a parametrization for this plane in rectangular coordinates is
r = < x, 0, z >. However, since you know that this plane intersects a cylinder, you may want to change to cylindrical coordinates. (Then again since you are integrating xy over this surface, you don't need calculate anything for this plane. Do you see why?)

What about the other plane?

This particular cylinder has its axis on the y-axis, rather than the z-axis, so y is an independent variable. If you use cylindrical coordinates, which of r or $\theta$
is constant in the parametrization of the cylinder?

 

[HallsofIvy]

Generally speaking "nice" function is differentiable. Any time a surface has sharp edges, as where the cylinder meets the planes, you cannot describe it with a single differentiable function. Any point in 3 space has "position vector",
$\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}$

Every point in the plane y= 0 has the obvious position vector $\vec{r}(x, z)= x\vec{i}+ z\vec{k}$. Every point in the plane x+ y= 2 satisifies y= 2- x so you can write its position vector $\vec{r}(x,z)= x\vec{i}+ (2- x)\vec{j}+ z\vec{k}$. Finally, since $cos^2(\theta)+ sin^2(\theta)= 1$, we can set $x= cos(\theta)$ and $z= sin(\theta)$ for the the cylinder $x^2+ z^2= 1$.

 

[flyingpig]

Ok thanks guys looks like I was just making iths more difficult than it needs to be trying to parametrize the whole thing