Evaluate the line integral

$$ \int_C zdx + xdy + xydz $$

where C is the path of the helix r(t) = (4cost)i + (4sint)j + (t)k on $0\le t \le 2\pi$

I solve this problem, but my answer was wrong.

x= 4cost

y= 4sint

z= t

dx= -4sint dt

dy= 4cost dt

dz= dt

I plugged these into the integral above and integrated to get

$$ 4tcost - 4sint + 4t +sin(4t) + 4sint \ |_0^{2\pi} $$

I solved this and got 16$\pi$ Can someone tell me where I am going wrong?

  • $\begingroup$ The primitive function looks wrong, but without giving more details on how you get there, I can only say it is in the step from simplifying $z\,dx+x\,dy+xy\,dz$ or doing the actual primitive it goes bad. I suggest you provide the details of that calculations if you need further help. $\endgroup$ – mickep Apr 4 '15 at 19:56

Your primitive is wrong. We have:

$$ \int_0^{2\pi}(16 \cos^2 t+16 \cos t \sin t-4t \sin t)dt= $$ $$ = 4\left(2t-\sin t +\sin 2t-2\cos^2 t+t \cos t \right) |_0^{2\pi}= 24 \pi $$


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