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?
Best Answer
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}