Example 9. Calculate the work of the force $\vec{F}=(x-y, 1)$ along the semicircle $x^{2}+y^{2}=4(y \geqslant 0)$ when a material point moves from $A(2,0)$ to $B(-2,0)$.
Solution. The work of force $F$ is calculated by the formula (see section 5.2) $A=\int_{L} \vec{F} \cdot \overrightarrow{d s}$, where $\vec{F}=(x-y, 1), \overrightarrow{d s}=(d x, d y)$. We will use the parametric equations of $L$ (Fig. 3.22): $x=2 \cos t, y=2 \sin t, t \in[0 ; \pi]$. In this case,

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-069.jpg?height=268&width=421&top_left_y=885&top_left_x=129)

Fig. 3.22

$$
d x=-2 \sin t d t, d y=2 \cos t d t
$$

$$
\begin{aligned}
& \vec{F} \cdot \overrightarrow{d s}=(x-y) d x+d y= \\
= & -(2 \cos t-2 \sin t) 2 \sin t d t+2 \cos t d t= \\
= & \left(4 \sin ^{2} t-4 \sin t \cos t+2 \cos t\right) d t=
\end{aligned}
$$

$$
=(2(1-\cos 2 t)-2 \sin 2 t+2 \cos t) d t
$$

Therefore, the work of force $\vec{F}$ is

$$
\begin{aligned}
& A=2 \int_{0}^{\pi}(1-\cos 2 t-\sin 2 t+\cos t) d t= \\
& \quad=\left.2\left(t-\frac{1}{2} \sin 2 t+\frac{1}{2} \cos 2 t+\sin t\right)\right|_{0} ^{\pi}=2 \pi
\end{aligned}
$$

Answer. $A=2 \pi$.