Question

\( \hat{i} \) and \( \hat{j} \) denote unit vectors in the \( x \) and \( y \) directions, respectively. The outward flux of the two-dimensional vector field \( \vec{v} = x \hat{i} + y \hat{j} \) over the unit circle centered at the origin is \_\_\_\_\_\_ (rounded off to two decimal places).

Hint:

To calculate the outward flux of a vector field through a closed curve, use the divergence theorem, which relates the flux to the divergence of the vector field over the enclosed area.

Answer 1

The outward flux of a vector field \( \vec{v} \) through a closed curve \( C \) is given by the surface integral: \[ \Phi = \oint_C \vec{v} \cdot \hat{n} \, ds \] where \( \hat{n} \) is the unit normal vector to the curve \( C \), and \( ds \) is the differential element of length along the curve. In this case, the vector field is \( \vec{v} = x \hat{i} + y \hat{j} \), and the curve is the unit circle centered at the origin. Since the vector field \( \vec{v} \) is the position vector itself, the flux through the unit circle is equivalent to the divergence of \( \vec{v} \) integrated over the area enclosed by the circle. The divergence of \( \vec{v} = x \hat{i} + y \hat{j} \) is: \[ {div}(\vec{v}) = \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (y) = 1 + 1 = 2 \] The flux is then the integral of the divergence over the area of the unit circle: \[ \Phi = \int_A 2 \, dA \] where \( A \) is the area of the unit circle. The area of the unit circle is \( \pi \), so: \[ \Phi = 2 \times \pi = 2\pi \] Thus, the outward flux is \( \Phi = 2\pi \). Rounded off to two decimal places: \[ \Phi \approx 6.28 \]