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Find the circulation and the flux of the field F = yi - xj around and across the following cines.

a. The circle r(t) = (cos t)i - (sin t)j, 0 <= t< = 2pi
b. The ellipse r(t)=(cost)i+(3 sin t)j, 0<= t <= 2pi

1. The circulation of F around the circle is ____________
2. The flux of F across the circle is____________


Sagot :

Explanation:

To solve for the circulation and flux of the vector field \( \mathbf{F} = y \mathbf{i} - x \mathbf{j} \) around and across the given curves, we'll use the following principles:

1. **Circulation of \(\mathbf{F}\) around a closed curve \(C\)** is given by:

\[

\oint_C \mathbf{F} \cdot d\mathbf{r}

\]

2. **Flux of \(\mathbf{F}\) across a closed curve \(C\)** is given by:

\[

\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS

\]

where \(S\) is the surface bounded by \(C\), and \(\mathbf{n}\) is the unit normal vector to \(S\).

### Part (a): Circle \( r(t) = \cos(t) \mathbf{i} - \sin(t) \mathbf{j} \), \( 0 \leq t \leq 2\pi \)

**1. Circulation:**

The parameterization of the circle is \( \mathbf{r}(t) = \cos(t) \mathbf{i} - \sin(t) \mathbf{j} \).

Compute \( d\mathbf{r} \):

\[

d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-\sin(t) \mathbf{i} - \cos(t) \mathbf{j}) dt

\]

The vector field \( \mathbf{F} \) in terms of \( t \) is:

\[

\mathbf{F} = y \mathbf{i} - x \mathbf{j} = (-\sin(t)) \mathbf{i} - (\cos(t)) \mathbf{j}

\]

Compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \):

\[

\mathbf{F} \cdot d\mathbf{r} = (-\sin(t) \mathbf{i} - \cos(t) \mathbf{j}) \cdot (-\sin(t) \mathbf{i} - \cos(t) \mathbf{j}) dt = (\sin^2(t) + \cos^2(t)) dt = dt

\]

Integrate over \( 0 \leq t \leq 2\pi \):

\[

\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} dt = 2\pi

\]

Thus, the circulation of \( \mathbf{F} \) around the circle is \( 2\pi \).

**2. Flux:**

Compute \( \nabla \times \mathbf{F} \):

\[

\nabla \times \mathbf{F} = \left( \frac{\partial (-x)}{\partial y} - \frac{\partial y}{\partial x} \right) \mathbf{k} = (-1 - 1) \mathbf{k} = -2 \mathbf{k}

\]

The area of the circle \( S \) is:

\[

\text{Area} = \pi \times r^2 = \pi \times 1^2 = \pi

\]

Since \( \mathbf{n} \) (the unit normal vector to the plane) is \( \mathbf{k} \), the flux through \( S \) is:

\[

\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \iint_S -2 \, dS = -2 \times \pi = -2\pi

\]

Thus, the flux of \( \mathbf{F} \) across the circle is \( -2\pi \).

### Summary

1. The circulation of \( \mathbf{F} \) around the circle is \( 2\pi \).

2. The flux of \( \mathbf{F} \) across the circle is \( -2\pi \).