Question

line integrals: a particle moves along an elliptic curve in a forcefield F. The ellips is in the xy plane and defined by (x/4)²+(y/3)²=1. F=(3x-4y+6z)e_x+(4x-2y+2z)e_y+(2xz-4y²-z^4) e_z. Determine the work it takes for a particle to move round the ellips once. TBC

Answer 1

From the ellipsis equation I conclude that

x=4 cos(t)
y=3 sin(t)

Work is
\[W \int\limits_{C}^{}F.dr\]

So, the integral would be the F given with x and y replaced with their t parameters. The integrals bounds are 0<t<2pi. The F equation in elementary vectors mentions z as well, but the movement is only in the xy plane. So, do I drop the third term in the F equation? Same for the z's mentioned in the first and second term as they would be 0?

Answer 2

I ended up having W=0 because sin(2pi) and sin (0) are both 0, and cos(2pi) and cos (0) are both 1 and subtracting 1-1=0

Answer 3

im getting \(\large 96 \pi\)

Answer 4

http://www.wolframalpha.com/input/?i=%5Cint_0%5E%7B2%5Cpi%7D+%28%283%284cost%29-4%283sint%29%29%28-4sint%29+%2B+%284%284cost%29-2%283sint%29%29%283cost%29%29dt

Answer 5

z = 0
dz = 0

one way to see that W cannot be 0 is by noticing that curl of given vector field is NOT 0, so work need not be 0 when you get back to your starting position

Answer 6

are you familair with green's theorem ?

Answer 7

thats another option to double confirm your work

Answer 8

But there are no dxs, dys and dzs in the given function. I was not sure what I needed to do with the vectors. I just used them as 1.
I'm just reading and studying that part (Green's for double integrals)

Answer 9

I meant the vectors ex, ey and ez.

Answer 10

\[\large \begin{align}\\ W &= \int\limits_{C}^{}F.dr \\&= \int\limits_{C}^{}\langle 3x-4y, ~4x-2y\rangle .\langle dx, dy\rangle \end{align}\]

Answer 11

thats what you get after plugging in z=0, dz=0, eh ?

Answer 12

yeah you should

Answer 13

I didn't write the dxs and dys, because there weren't any written in the F equation

Answer 14

you may think dr as vector <dx, dy>

Answer 15

your parameterization looks good,
just apply it and evaluate the integral

Answer 16

ah ok, you're right

Answer 17

I got 96 pi now too

Answer 18

good, lets work it using green's theorem ?

Answer 19

ok sure

Answer 20

what does green's theorem say ?

Answer 21

not sure yet, I was just starting to read that part, after getting the basics of doing double integrals

Answer 22

\[\large \oint_C F.dr = \iint curl(F) dxdy\]

Answer 23

it would mean that I can write a circular integral as a double integral?

Answer 24

work done along a CLOSED LOOP equals the double integral for the area bounded by that LOOP.

Answer 25

work done along a CLOSED LOOP equals the double integral for the area bounded by that LOOP of Cirl of the vector field.

i think the integral symbols make more sense than words :P