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Find the work done by the force field $F(x, y) = \langle 2x \sin(y), 2y \rangle$ on a particle that moves along the parabola $y = x^2$ from $(-1, 1)$ to $(2, 4)$.

So to use line integrals to solve this I took $r(x) = \langle x, x^2 \rangle$. Then $$\int_{-1}^2 \langle 2x \sin{x^2}, 2x^2 \rangle \cdot \langle 1, 2x \rangle = \int_{-1}^2 2x \sin{x^2} + 4x^2 dx.$$ Does that work?

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    $\begingroup$ That looks fine, except the integrand should have $4x^3$, and not $4x^2$ I believe. $\endgroup$ – msteve Jul 27 '14 at 22:36
  • $\begingroup$ @msteve oh yeah, thanks! $\endgroup$ – Shaurya Dhingra Jul 27 '14 at 22:37
  • $\begingroup$ @msteve You should make that an answer. $\endgroup$ – Mark Fantini Jul 27 '14 at 23:34
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The integral should be the following: \begin{align} \int_{-1}^2 F(x,x^2) \cdot r(x) dx &= \int_{-1}^2 \langle 2x\sin(x^2), 2x^2 \rangle \cdot \langle 1, 2x \rangle dx \\ &= \int_{-1}^2 (2x\sin(x^2) + 4x^3) dx \\ &= -\cos(x^2) + x^4 \bigg|_{-1}^2 \\ &= 15 + \cos(1) - \cos(4). \end{align}

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