Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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 \( -2\pi \).
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.