The work done by the force $F = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors in $\vec{OX}$ and $\vec{OY}$ directions, respectively, along the upper half of the circle $x^2 + y^2 = 1$ from $(1,0)$ to $(-1,0)$ in the $xy$-plane is
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} $.
Important Questions from Area Under Curve
- Let $I$ be the integral defined as follows: $$I = \int_{0}^{1} \int_{0}^{\sqrt{y}} dx dy + \int_{1}^{2} \int_{\sqrt{y-1}}^{1} dx dy$$ If the order of the integration is changed, then which one of the following is the correct expression for $I$?
Let $\alpha = \iint_S \vec{F} \cdot \hat{n} \, dS$, where $\vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k}$ and $S$ is the sphere with centre at $(3, -1, 2)$ and radius 9. Here, $\hat{n}$ is the unit normal drawn outward and $\hat{i}, \hat{j}, \hat{k}$ are unit vectors.
Then the value of $\frac{1}{36\pi} \alpha$ is equal to ________. (answer in integer)
The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).
If the line $y = \alpha x$, $\alpha \geq \sqrt{2}$, divides the area of the region
$R: = \{(x, y) \in \mathbb{R}^2| 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2\}$
into two equal parts, then the value of $\alpha$ is equal toLet $C: x^2 + y^2 =9$ be the circle in $R^2$ oriented positively. Then $\frac{1}{\pi}\oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy$ equals_____.
