Question

I'm doing a Stokes Theorem problem now, trying to learn/remember all of this stuff quickly. Workings so far posted below momentarily.

Answer 1

http://i.imgur.com/DFNRdyG.png

Answer 2

ok

Answer 3

Now just about to calculate \[\triangledown \times F\cdot n \]

Answer 4

Is n just....n is just also one? Just the k hat vector?

Answer 5

So \[Curl \ F \cdot n = 2 \cdot 1 = 2\]

Answer 6

So the integrand is two?

Answer 7

yes

Answer 8

normal line for something in the x y plane points straight up or down in z axis

Answer 9

and nhat is unit normal so, <0,0,-1> or <0,0,1> is a normal line, depending on of u wanna go clockwise or ccw

Answer 10

Alright, so now the bounds of integration are established by the given curve. Yeah. Going to try to set up the integral now.

Answer 11

ok

Answer 12

u shud use a transformation

Answer 13

make the ellipse into a circle

Answer 14

and do polar

Answer 15

and use jacobian to do transform i know ellipse to circle is straight foward but.. practice some jacobians

Answer 16

Alright, lemme take a shot at the transform to get the circle in the right shape, I vaguely remember how to do it, going to try now.

Answer 17

ok

Answer 18

\[x = 2u, \ \ \ y = v\]

Answer 19

yeah

Answer 20

Wait, is the 2 supposed to be in the denominator, e.g. 2/u? One second.

Answer 21

basially u wanna get into a circle so
4x^2+y^2=4
a^2+b^2=4
so
a^2=4x^2
a=2x
b^2=y^2
b=y

Answer 22

\[u^2 + v^2 = 4; \ \ \ r = 2\]

Answer 23

I need to also take the Jacobian to figure out whatever the integrand is supposed to be multiplied by, trying to remember how to do that properly, heh

Answer 24

Be back in like 2 minutes, dealing with something IRL

Answer 25

ok

Answer 26

jacobian comes from cross product to get area

Answer 27

Could you elaborate on that? I've always seen the jacobian as the determinant of a matrix with partial derivatives involved.

Answer 28

@Jhannybean , could you help me set up this Jacobian? I've forgotten like, half of the stuff on this after putting all my time into ODE...re-reading on it now.

Answer 29

And do I want\[J(u,v)\]or\[J(x,y)?\]

Answer 30

(Trying to figure out how to set up a 2x2 determinant matrix in LaTeX atm.)

Answer 31

Arg...Jacobians, gotta look this up in my book.

Answer 32

I know/remember how to do this, but now I'm just trying to figure out how to put it into LaTeX, lul. I'll write out what I'm trying to do "by hand".

Answer 33

http://i.imgur.com/IupsZ0n.png

Answer 34

ohh,
`\begin{bmatrix}#&#\\#&#
\end{bmatrix}`

Answer 35

Just insert your numbers where the pound signs are.

Answer 36

Alright, cool.

I just wrote out a whole page of stuff on my tablet and the program crashed and I lost all of it, the disadvantage of "writing things out" ;_;

Answer 37

\[\begin{bmatrix}2&0\\0&1 \end{bmatrix}\]

Answer 38

WOO!

Answer 39

...I need to seriously start on my own calc homework -___- Gotta get HW done and then start reviewing Stokes Theorem since we're beginning to learn it this week!!

Answer 40

\[\frac {\partial x}{\partial u} = 2 \ du = 2, \ \ \ \frac{\partial y }{\partial v} = du = 1\]

Alright, so the integrand, dealing with the Jacobian of the transformation, and the \[\triangledown \times F \cdot n = 2,\]

Answer 41

"And then start reviewing", lol, you're totally on the ball/ahead, congratulations.

Answer 42

\[\int\limits_{0}^{2 \pi}\int\limits_{0}^{2}4 r \ dr \ d \theta\]

Answer 43

(Hesitating on that, looks too simple)

Answer 44

Which equals 16pi if I did things right.

Answer 45

it is much simpler than that, you could have avoided Jacobian stuff altogether because the change of variables is linear and uniform

Answer 46

(wot m8, please elabor8)

Answer 47

(...ganeshie8)

Answer 48

\[x = 2u, \ \ \ y = v\]

Notice that you're scaling uniformly in x direction by a factor of 2 and in y direction bby a factor of 1. This scaling is same through out the graph. So the area also scales same way :


\[dx = 2du, \ \ \ y = dv \implies dxdy = 2dudv\]

Answer 49

Oh, okay, I thought you meant some wholly alternate method, not just shortcutting the Jacobian, but yeah, that makes sense

Answer 50

No, I only want to point out that you can avoid that determinant in certain cases. For example, you can avoid it for change of variables of form : x= mu, y = nu

dx = mdu
dy = ndv

the scale factor is simply mn. you don't need to find partials and carry out the process to see this..

Answer 51

but im sure your prof will cut marks if u dont show him jacobian work

Answer 52

so you can't use this in your hw problems where your prof expects to see jacobian

Answer 53

Yuh, well, he never has us turn in homework anyways, which I think is incredibly lame, he just has suggested problems and online, computer-graded stuff.

Answer 54

But yeah, 16 pi, right?

Answer 55

Supposed to be 4pi, nevermind, I made a mistake somewhere.

Answer 56

yeah in change of variables i think

Answer 57

try \[x = u/2, ~~y = v\]

Answer 58

Ayyy, I suggested that, too, but I got shot down, lol. I'll try that.

Answer 59

use the earlier trick : dxdy = 1/2 dudv

Answer 60

So the Jacobian in this case would just be\[J(u,v)=\frac{1}{2},\]the integrand would be identical, so you would be left with-yep

Answer 61

The integrand would be one instead of four, which would scale by four identically like the other integral did by four.

Answer 62

Alright, yeah. Makes sense. Thank you!

Answer 63

Looks good!