Consider the vector field $ \mathbf{F} =\Big(-\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\Big) $ and the curve defined by $ \frac{x^2}{9} + \frac{y^2}{1} + z = 2 $. Calculate the work done by the force $ \mathbf{F} $ to move a particle along the curve $ \mathbf{r} $ . I want to calculate it without using Stokes' Theorem.
We ca calculate it using:
$$ W = \int_C \mathbf{F} \cdot d\mathbf{r} $$
The position vector $ \mathbf{r} $ for the given curve is parameterized as follows:
$$ \mathbf{r}(t) = \langle 3\cos(t), \sin(t), 2 - \frac{8}{9}\cos^2(t) - \frac{2}{9}\sin^2(t) \rangle $$
The parameter $ t $ ranges from $ 0 $ to $ 2\pi $.
We have that,
$$ d\mathbf{r} = \langle -3\sin(t), \cos(t), \frac{16}{9}\cos(t)\sin(t) \rangle \, dt $$
$$ \mathbf{F}(\mathbf{r}) = \Bigg(-\frac{\sin(t)}{9\cos^2(t) + \sin^2(t)}, \frac{3\cos(t)}{9\cos^2(t) + \sin^2(t)}\Bigg) $$
Now, the dot product $ \mathbf{F} \cdot d\mathbf{r} $ is given by:
$$ \mathbf{F} \cdot d\mathbf{r} = -\frac{\sin(t)}{9\cos^2(t) + \sin^2(t)} \cdot (-3\sin(t)) + \frac{3\cos(t)}{9\cos^2(t) + \sin^2(t)} \cdot \cos(t) + 0 $$
Integrating:
$$ W = \int_0^{2\pi} \left(3\sin^2(t) + 3\cos^2(t)\right) \, dt $$
This simplifies to:
$$ W = \int_0^{2\pi} 3 \, dt = 3 \cdot [t]_0^{2\pi} = 6\pi $$
So, the work done by the force field $ \mathbf{F} $ to move a particle along the given curve is $ 6\pi $.
Could you please tell me if my solution is correct?