- #1
bigevil
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Homework Statement
For the field [tex]\bold{F} = (y+z) \bold{i} - (x+z) \bold{j} + (x+y) \bold{k}[/tex] find the work done in moving a particle around the following closed curve:
from the origin to (0,0,2π) on the curve x=1-cos t, y=sin t, z=t; and back to the origin along the z-axis. The answer is 2π. (This question is from Mary L Boas' textbook btw, which is why I have the answer.)
Homework Equations
We get:
dx = sin t dt
dy = cos t dt
dz = dt
x = 1 - cos t
y = sin t
z = t
I did this two ways and didn't get either right. Could someone point out what was wrong with both methods and give me a hint?
First method: I assumed that dx = dy = 0. This gives [tex]\int_0^{2\pi} (1-cos t + sin t) dt = 2\pi[/tex]. But going by this method, going back to origin means that the integral for the reverse action is -2π, so the work is 0. This method is probably wrong because, obviously, x=1 - cos t is not a "static" function.
Second method: Expand everything in full, which means
[tex]W = \int F . dr = \int (sin t + t)(sin t) dt - (1 - cos t + t)(cos t) dt + (1 - cos t + sin t) dt = \int_0^{2\pi} 2 + sin t - 2 cos t dt = 4\pi[/tex], which is also wrong. For the reverse action, the limits are flipped and I get 0 again.
I think this is the right method, but there's something wrong with the integration here although its supposed to be pretty simple.