Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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 trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
