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Evaluate the line integral $\int_C \mathbf F \cdot \, \mathrm d \mathbf r$ where, $\mathbf F = x \mathbf i+y \mathbf j + xy \mathbf k$ and C is parameterised by $\mathbf r(t)= \cos t \mathbf i+\sin t \mathbf j + t \mathbf k,t\in[0,\pi]$

Using $$\int_C \mathbf F \cdot \, \mathrm d \mathbf r = \int^b_a \mathbf F (\mathbf r(t))\mathbf r'(t) dt $$

I have begun to answer the question, this is my current unfinished solution:

$$=\int^\pi_0 (\cos t\mathbf i + \sin t\mathbf j +\cos t \sin t\mathbf k)(-\sin t\mathbf i + \cos t\mathbf j + \mathbf k) dt$$

$$=\int^\pi_0(-\sin t \cos t+ \sin t \cos t+\sin t \cos t) dt$$

$$=\int^\pi_0(\sin t \cos t) dt$$

Not sure if this is correct up until this point.

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  • $\begingroup$ Looks good so far. $\endgroup$ – Matthew Leingang Jan 2 '18 at 18:06
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This has been correct so far, well done!

From here, you can integrate by parts to obtain the final answer.

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  • $\begingroup$ Simplifying the integrand via a double-angle identity is easier here, I think. $\endgroup$ – amd Jan 3 '18 at 2:08
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As noted by the other users your solution up to that point is right. To finish it off make use of the indentity $\cos t \cdot \sin t = \frac{\sin(2t)}{2}$. To get:

$$\int^\pi_0(\sin t \cos t) dt = \int_0^{\pi}\frac{\sin(2t)}{2}dt = \frac{-\cos(2t)}{4} \bigg |_{t=0}^{t=\pi} = 0$$

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