Work Done Calculation Using Line Integral

The work done ($W$) by a force vector $\vec{F}$ along a path $C$ is calculated using the line integral:

$W = \int_C \vec{F} \cdot d\vec{r}$

Here, the force is given by $\vec{F} = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, and the path $C$ is the upper half of the circle $x^2 + y^2 = 1$ traversed from $(1,0)$ to $(-1,0)$.

Parametrizing the Path

We parametrize the upper half of the unit circle using the angle $\theta$:

• $x = \cos\theta$
• $y = \sin\theta$

The path starts at $(1,0)$, which corresponds to $\theta = 0$. The path ends at $(-1,0)$, which corresponds to $\theta = \pi$. Thus, $\theta$ ranges from $0$ to $\pi$.

The differential displacement vector is $d\vec{r} = dx \hat{i} + dy \hat{j}$. Differentiating the parametric equations:

• $dx = -\sin\theta \, d\theta$
• $dy = \cos\theta \, d\theta$

So, $d\vec{r} = (-\sin\theta \, d\theta)\hat{i} + (\cos\theta \, d\theta)\hat{j}$.

Expressing Force and Calculating Dot Product

Substitute the parametric equations into the force vector $\vec{F}$:

• $x + y = \cos\theta + \sin\theta$
• $x^2 + y^2 = (\cos\theta)^2 + (\sin\theta)^2 = 1$

Therefore, $\vec{F} = (\cos\theta + \sin\theta)\hat{i} - (1)\hat{j}$.

Now, calculate the dot product $\vec{F} \cdot d\vec{r}$:

$ \vec{F} \cdot d\vec{r} = [(\cos\theta + \sin\theta)\hat{i} - \hat{j}] \cdot [(-\sin\theta \, d\theta)\hat{i} + (\cos\theta \, d\theta)\hat{j}] $

$ \vec{F} \cdot d\vec{r} = (\cos\theta + \sin\theta)(-\sin\theta) \, d\theta + (-1)(\cos\theta) \, d\theta $

$ \vec{F} \cdot d\vec{r} = (-\sin\theta \cos\theta - \sin^2\theta - \cos\theta) \, d\theta $

Evaluating the Integral

Set up the definite integral for work done:

$W = \int_{0}^{\pi} (-\sin\theta \cos\theta - \sin^2\theta - \cos\theta) \, d\theta$

Evaluate each term separately:

• $ \int_{0}^{\pi} -\sin\theta \cos\theta \, d\theta = [-\frac{\sin^2\theta}{2}]_{0}^{\pi} = 0 - 0 = 0 $
• $ \int_{0}^{\pi} -\sin^2\theta \, d\theta = \int_{0}^{\pi} -\frac{1 - \cos(2\theta)}{2} \, d\theta = [-\frac{1}{2}(\theta - \frac{\sin(2\theta)}{2})]_{0}^{\pi} = -\frac{1}{2}(\pi - 0) - (-\frac{1}{2}(0 - 0)) = -\frac{\pi}{2} $
• $ \int_{0}^{\pi} -\cos\theta \, d\theta = [-\sin\theta]_{0}^{\pi} = -\sin(\pi) - (-\sin(0)) = 0 - 0 = 0 $

Summing the results:

$ W = 0 - \frac{\pi}{2} + 0 = -\frac{\pi}{2} $

The work done is $ -\frac{\pi}{2} $.