IDNLearn.com offers a unique blend of expert answers and community-driven knowledge. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
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 \( -2\pi \).
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.
