Question

Consider a vector \(\mathbf{F} = \frac{1}{\pi}[-\sin y \hat{i} + x(1 - \cos y)\hat{j}]\). The value of the integral \(\oint \mathbf{F} \cdot d\mathbf{r}\) over a circle \(x^2 + y^2 = 1\) evaluated in the anti-clockwise direction is _____. (in integer)

Hint:

Whenever you see a line integral over a closed loop (\(\oint\)), especially in 2D, consider using Stokes' Theorem (or Green's Theorem, which is the 2D version). It often simplifies the problem from a potentially complicated line integral to a much simpler surface integral, particularly if the curl of the vector field is a constant or a simple function.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
This problem asks for the evaluation of a line integral of a vector field over a closed loop (a circle). A convenient method for solving such problems is to use Stokes' Theorem, which relates the line integral over a closed loop to the surface integral of the curl of the vector field over the surface enclosed by the loop.
Step 2: Key Formula or Approach:
Stokes' Theorem states: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \] where C is the closed curve (the circle \(x^2 + y^2 = 1\)) and S is the surface enclosed by C (the disk \(x^2 + y^2 \le 1\)).
The vector field is \(\mathbf{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\) with \(F_x = -\frac{1}{\pi}\sin y\), \(F_y = \frac{x}{\pi}(1 - \cos y)\), and \(F_z = 0\). The curl is given by \(\nabla \times \mathbf{F} = \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{k}\).
Step 3: Detailed Explanation:
First, calculate the components of the curl: \[ \frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x} \left[\frac{x}{\pi}(1 - \cos y)\right] = \frac{1}{\pi}(1 - \cos y) \] \[ \frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y} \left[-\frac{1}{\pi}\sin y\right] = -\frac{1}{\pi}\cos y \] Now, calculate the curl: \[ \nabla \times \mathbf{F} = \left[ \frac{1}{\pi}(1 - \cos y) - \left(-\frac{1}{\pi}\cos y\right) \right] \hat{k} \] \[ \nabla \times \mathbf{F} = \left[ \frac{1}{\pi} - \frac{1}{\pi}\cos y + \frac{1}{\pi}\cos y \right] \hat{k} = \frac{1}{\pi}\hat{k} \] Next, we set up the surface integral. The surface S is the unit disk in the x-y plane. The differential surface element \(d\mathbf{S}\) is \(dx dy \hat{k}\) for an anti-clockwise path. \[ \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S \left(\frac{1}{\pi}\hat{k}\right) \cdot (dx dy \hat{k}) = \iint_S \frac{1}{\pi} dx dy \] We can take the constant \(1/\pi\) out of the integral: \[ \frac{1}{\pi} \iint_S dx dy \] The integral \(\iint_S dx dy\) is simply the area of the surface S, which is the area of a circle with radius \(r=1\). Area of S = \(\pi r^2 = \pi (1)^2 = \pi\). Finally, substitute the area back into the expression: \[ \oint \mathbf{F} \cdot d\mathbf{r} = \frac{1}{\pi} \times (\text{Area of S}) = \frac{1}{\pi} \times \pi = 1 \] Step 4: Final Answer:
The value of the integral is 1.