# stokes

• May 7th 2010, 01:15 PM
alexandrabel90
stokes
find the line integral of F. dr where C is the perimeter of a rectangle formed by connected (0,2,0) ,(0,0,2), (2,0,2) and (2,2,0) where F is the vector field F = (z^2, x^2, y^2)?

in order to use stokes theorem, i tried to find the normal vector by using 2 equations on the plane and i got N= (0,4,4). then the unit normal will be n=1/
√2 (0,1,1).

however if i form the equation of the plane, i get z= 2-y and hence the normal will be (0,1,1) instead.

may i know what is the normal vector that i have to use for stokes theorem?
• May 7th 2010, 04:23 PM
zzzoak
If you make a unit vector from (0,1,1) you get the same as first one.
• May 8th 2010, 02:45 AM
alexandrabel90
but if i am using a normal (non unit) vector, then wouldnt (0,4,4) and (0,1,1) matter to the product? becos in stoke theorem, i thought the magnitude of the normal vector will be cancelled out.
• May 8th 2010, 03:30 AM
HallsofIvy
Yes, that's why it doesn't matter! (0, 1, 1), (0, 4, 4), and $\sqrt{2}(0, 1, 1)$ all point in the same direction and are normal to the same plane. The integral in Stoke's theorem is often expressed as $\int\int \nabla\times \vec{f} \vec{n}dS$ where $\vec{n}$ is the unit normal vector.

However, that's a bit misleading- and may be the source of your confusion. If a surface is given by z= f(x,y), then a standard formula for dS involves $\sqrt{1+ \left(\frac{\partial z}{\partial x}\right)^2+ \left(\frac{\partial z}{\partial x}\right)^2}$ which is, really, a length of a normal curve. That is " $\vec{n}dS$" involves dividing by the length of a normal vector to get $\vec{n}$ and multiplying that length back in to get dS! In fact, I prefer to write " $d\vec{S}$" incorporating $\vec{n}$ into dS- it is the vector normal to the surface whose length is the "differential of surface area".

And that is relatively easy to calculate- write the surface in terms of the "position vector" of each point in terms of two parameters. In the case where z= f(x,y) you can take x and y themselves as parameters: $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}$.

In the case of this plane, we can write the equation of the plane by taking the cross product of the vectors (0, 0, 2)- (0, 2, 0)= (0, -2, 2) and (2, 2, 0)- (0, 2, 0)= (2, 0, 0) that form two consecutive sides of the rectangle. That cross product is either (0, 4, 4) or (0, -4, -4) depending on the order in which you take the cross product. That tells us that the equation of the plane is either -4(y-0)- 4(z- 2)= 0 or 4(y-0)+ 4(z- 2)= 0. Those both reduce to the same equation, 4y+ 4z= 8, or z= 2- y, or course. And that we can write, as a "position vector" of two parameters, as $\vec{r}(x,y)= x\vec{i}+ y\vec{j1}+ (2- y)\vec{k}$

Now, given a surface written as a "position vector" of two parameters, $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$ the two partial derivatives, $\vec{r}_u= x_u\vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are vectors in the tangent surface to the sufrace at each point and their cross product, the "fundamental vector product" for the surface, gives the "differential of surface area". That is, the "scalar differential of surface area", dS, is $||\vec{r}_u\times\vec{r}_v||dudv$ and the "vector differerential of surface area", $\vec{n}dS= d\vec{S}$ is simply $\vec{r}_u\times\vec{r}_v dudv$.

Notice that there is an ambiguity in that- the cross product is "skew-commutative" and $\vec{r}_u\times\vec{r}_v= -\vec{r}_v\times\vec{r}_u$. Which order do we use? That depends on the orientation of the path and surface.

If we write this plane as $\vec{r}(x,z)= x\vec{i}+ y\vec{j}+ (2- y)\vec{k}$, we have $\vec{r}_x= \vec{i}$ and $\vec{r}_y= \vec{j}- \vec{i}$. The cross product, in the order $\vec{r}_x\times\vec{r}_y$,is $\vec{j}+ \vec{k}$ and,in the order $\vec{r}_v\times\vec{r}_u$ is $-vec{j}- \vec{k}$.

The "vector differential or surface area", $d\vec{S}= \vec{n}dS$ is either (0, 1, 1)dxdy or (0, -1, -1)dxdy.

And that difference is important- there is a slight ambiguity in this problem itself- you haven't given an orientation for the integrals. That is, saying "find the perimeter of a rectangle formed by connected (0,2,0) ,(0,0,2), (2,0,2) and (2,2,0)" doesn't say in which direction to integrate! Are we to integrate from (0, 2, 0) to (2, 2, 0) to (2, 0, 2) to (0, 0, 2) and then back to (0, 2, 0) or around that same loop in the opposite direction? The two integrals will differ in sign, of course, exactly like (0, 1, 1) and (0, -1, -1) differ.

The rule in Stoke's theorem is sort of a "right hand rule". Imagine that you are walking around the bounding path in such a way that you are facing in the direction of integration on the path and your right hand extends inward. Then your head is in the direction of the normal vector for that orientation.

Here, if we are going from (0, 2, 0) to (2, 2, 0) with right hand inside the rectangle, our head is pointing in the direction of positive y and z: the correct normal vector is (0, 4, 4) and $d\vec{S}= (0, 1, 1)dxdy$.

But if we are walking from (0, 2, 0) to (0, 0, 2)with right hand inside the rectangle, our head is pointing in the direction of negative y and z: the correct normal vector is (0, -4, -4) and $d\vec{S}= (0, -1, -1)dx dy$.