Question

Consider a unit circle C in the xy plane with center at the origin. The line integral of the vector field, $ \overrightarrow{F}(x, y, z) = -2y-3z\hat{y} + x\hat{z}$, taken anticlockwise over C is _________ π.

Answer 1

Correct Answer: 2

1. Understand the Vector Field and Geometry

Curve $C$: A unit circle in the $xy$-plane ($z=0$) centered at the origin.

Vector Field: The expression given is $\vec{F} = -2y - 3z\hat{y} + x\hat{z}$. In standard vector notation, the first term without a specific unit vector corresponds to the $\hat{i}$ (or $\hat{x}$) component. Thus, the vector field is:

$$\vec{F} = -2y\hat{i} - 3z\hat{j} + x\hat{k}$$

2. Method 1: Direct Parameterization

We can solve the line integral $\oint_C \vec{F} \cdot d\vec{r}$ by parametrizing the curve $C$.

Parameterization of the unit circle (anticlockwise):

$$x = \cos \theta$$

$$y = \sin \theta$$

$$z = 0$$

where $\theta$ varies from $0$ to $2\pi$.

Differential displacement vector ($d\vec{r}$):

$$dx = -\sin \theta \, d\theta$$

$$dy = \cos \theta \, d\theta$$

$$dz = 0$$

$$d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k} = (-\sin \theta \, d\theta)\hat{i} + (\cos \theta \, d\theta)\hat{j}$$

Evaluate $\vec{F}$ on the curve:

Since $z=0$ on the curve, the vector field simplifies to:

$$\vec{F} = -2(\sin \theta)\hat{i} - 3(0)\hat{j} + (\cos \theta)\hat{k}$$

$$\vec{F} = -2\sin \theta \, \hat{i} + \cos \theta \, \hat{k}$$

Calculate the dot product $\vec{F} \cdot d\vec{r}$:

$$\vec{F} \cdot d\vec{r} = (-2\sin \theta)(-\sin \theta \, d\theta) + (0)(\cos \theta \, d\theta) + (\cos \theta)(0)$$

$$\vec{F} \cdot d\vec{r} = 2\sin^2 \theta \, d\theta$$

Integrate:

$$I = \int_{0}^{2\pi} 2\sin^2 \theta \, d\theta$$

Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:

$$I = \int_{0}^{2\pi} 2 \left( \frac{1 - \cos 2\theta}{2} \right) d\theta$$

$$I = \int_{0}^{2\pi} (1 - \cos 2\theta) \, d\theta$$

$$I = \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{2\pi}$$

$$I = (2\pi - 0) - (0 - 0) = 2\pi$$

3. Method 2: Stokes' Theorem

Stokes' Theorem relates the line integral to a surface integral: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$.

Surface $S$: The unit disk in the $xy$-plane bounded by $C$.

Normal Vector $\hat{n}$: Since $C$ is anticlockwise in the $xy$-plane, $\hat{n} = \hat{k}$.

Curl of $\vec{F}$:

$$\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -2y & -3z & x \end{vmatrix}$$

The $\hat{k}$-component of the curl is:

$$\left( \frac{\partial (-3z)}{\partial x} - \frac{\partial (-2y)}{\partial y} \right)\hat{k} = (0 - (-2))\hat{k} = 2\hat{k}$$

Surface Integral:

$$\iint_S (2\hat{k}) \cdot \hat{k} \, dS = \iint_S 2 \, dS = 2 \times (\text{Area of unit circle})$$

$$I = 2 \times (\pi (1)^2) = 2\pi$$

Final Answer

The value of the line integral is $2\pi$.

The blank should be filled with 2.