# Thread: Two-form integral of plane in first octant

1. ## Two-form integral of plane in first octant

Let S be that portion of the plane $\displaystyle x + 2y + 2z = 4$ lying in the first octant, oriented with outward normal pointing upward.

Find $\displaystyle \int_{S}{zdx\wedge dy + ydz\wedge dx + xdy\wedge dz}$.

This is my first time trying to do a problem with a two-form, and I'm not really sure where to begin. Also, the book is not very helpful with understanding the wedge notation.

2. Originally Posted by Redding1234
Let S be that portion of the plane $\displaystyle x + 2y + 2z = 4$ lying in the first octant, oriented with outward normal pointing upward.

Find $\displaystyle \int_{S}{zdx\wedge dy + ydz\wedge dx + xdy\wedge dz}$.

This is my first time trying to do a problem with a two-form, and I'm not really sure where to begin. Also, the book is not very helpful with understanding the wedge notation.
The wedge notation is not really different from just "dxdy" except that it contains orientation information: $\displaystyle dx\wedge dy= -dy \wedge dx$ and, here, you are told how the surface is oriented.

With plane x+ 2y+ 2z= 4, we can write x= 4- 2y- 2z and so a position vector for a point on the plane is given by $\displaystyle \vec{r}= x\vec{i}+ y\vec{h}+ z\vec{k}= (4- 2y- 2z)\vec{i}+ y\vec{j}+ z\vec{k}$. The two derivatives, $\displaystyle \vec{r}_y= -2\vec{i}+ \vec{j}$ and $\displaystyle \vec{r}_z= -2\vec{j}+ \vec{k}$. The differential, in terms of the parameters y, z, is given by their cross product: $\displaystyle (\vec{i}- 2\vec{j}+ 2\vec{k})dydz$. Of course, I have chosen the order of vectors in of the product to give a positive $\displaystyle \vec{k}$ component "oriented with outward normal point upward".

Take the dot product of that with the given vector, $\displaystyle z\vec{i}+ y\vec{j}+ x\vec{k}= z\vec{i}+ y\vec{j}+ (4- 2y- 2z)\vec{k}$ and integrate. Projecting into the yz- plane, 2y+ 2z= 4 so y goes from 0 to 2 and, for each y, z goes from 0 to 2- y.

3. Thanks, that was very helpful!

It doesn't matter which variable you solve for to begin the problem, right?

I tried the whole problem over again by initially solving for y and finding $\displaystyle r_x$ and $\displaystyle r_x$ (so the integral ended up in terms of x and z), and the answer came out the same (after I altered the cross product so that the $\displaystyle k$ component was positive). Your method was a little easier computationally though.