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Evaluate the line integral zdx+xdy+xydz where is the path of the helix r(t)=(2cost)i+(2sint)j+(1t)k on 0 less than t less than 2

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make variable change to cyulindrical coordinates first with r=2
x=2cost y=2sint z=z so: dx=-2sint*dt dy=2cost*dt dz=dz
i got to \[\int\limits_{0}^{2\pi} -2 tsintdt +4\cos^2tdt + 4costsintdt\] but i cant figure put how to integrate

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Other answers:

t is bound above by 2 or 2pi?
first term: integrate by parts second term: double angle formula \(\cos^2t=\frac12(1+\cos(2t))\) third term: u=sin(t)
\[0\le t \le 2\pi\]
that makes more sense obviously so did you try it yet?
no the homework is not due till tonight so im studying for my test and will try it after!
good luck!
i got 3 pi but it was incorrect.
\[\int_0^{2\pi}-2t\sin t+4\cos^2t+4\sin t\cos tdt\]\[2t\cos t-2\sin t+2t+\sin(2t)+2\sin t|_0^{2\pi}\]is that what you got?
no my signs were messed up.
let me check with wolfram...
thats correct! i finished it!
i had too many subtraction signs.
cool :) yeah, the algebra can be the worst in calc 3
hahhaa. yeah. thanks so much. :)

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