rewrite hte surface integral as a lien integral

To rewrite the surface integral as a line integral, we will use Stokes' Theorem, which states:

\[
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}
\]

### Given:
- The vector field \( \mathbf{F} = (x - y)\mathbf{i} + (y - z)\mathbf{j} + (z - x)\mathbf{k} \).
- The surface \( S \) is the disk \( x^2 + y^2 = 4 \) at \( z = 1 \), and the curve \( C \) is the boundary of this disk.

### Step 1: Compute the Curl of \( \mathbf{F} \)
We already computed the curl:
\[
\nabla \times \mathbf{F} = -\mathbf{i} - \mathbf{j} - \mathbf{k}
\]

### Step 2: Surface Integral Setup
The surface integral is:
\[
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}
\]

### Step 3: Use Stokes' Theorem
Stokes' Theorem allows us to rewrite this surface integral as a line integral around the boundary curve \( C \):

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

### Step 4: Express the Line Integral
The curve \( C \) is the circle \( x^2 + y^2 = 4 \) in the plane \( z = 1 \).

In parametric form, the curve can be represented as:
\[
\mathbf{r}(t) = (2\cos t)\mathbf{i} + (2\sin t)\mathbf{j} + \mathbf{k}, \quad 0 \leq t \leq 2\pi
\]
Here, \( t \) is the parameter.

### Step 5: Find \( d\mathbf{r} \)
The differential \( d\mathbf{r} \) is:
\[
d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-2\sin t)\mathbf{i} + (2\cos t)\mathbf{j} \, dt
\]

### Step 6: Substitute into the Line Integral
Now substitute \( \mathbf{F} \) and \( d\mathbf{r} \) into the line integral:
\[
\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt
\]
Where \( \mathbf{F}(\mathbf{r}(t)) \) is \( \mathbf{F} \) evaluated at \( \mathbf{r}(t) = (2\cos t, 2\sin t, 1) \):
\[
\mathbf{F}(t) = (2\cos t - 2\sin t)\mathbf{i} + (2\sin t - 1)\mathbf{j} + (1 - 2\cos t)\mathbf{k}
\]

So the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \) becomes:
\[
\left[(2\cos t - 2\sin t)(-2\sin t) + (2\sin t - 1)(2\cos t)\right] dt
\]

This expression simplifies to:
\[
\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left[-4\cos t \sin t + 4\sin t \cos t - 2\sin t\right] dt
\]

This integral can now be computed to evaluate the line integral, completing the conversion from a surface integral to a line integral using Stokes' Theorem.

Would you like any more details or assistance in solving the integral?

### 8 Related Questions:
1. How do you find the parametric equations for a curve given its geometric description?
2. How does the orientation of the curve affect the outcome of the line integral?
3. What would change if the vector field \( \mathbf{F} \) were different?
4. Why is it necessary to compute the curl of the vector field in Stokes' Theorem?
5. Can Stokes' Theorem be applied to any surface with a given boundary?
6. What is the physical interpretation of a line integral in this context?
7. How do you handle cases where the curve \( C \) is not simple (e.g., has multiple components)?
8. What would the integral look like if the surface were parameterized differently?

### Tip:
When parameterizing curves, ensure the parameterization covers the entire curve exactly once and maintains the correct orientation.

Learn how to rewrite a surface integral as a line integral using Stokes' Theorem. This problem involves converting a given vector field and surface into a line integral over its boundary curve. Detailed step-by-step solution provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation