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Given the problem, we need to show that the area of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is equal to the magnitude of the work done by the force field \( \mathbf{F}(x, y) = -\frac{1}{2}y \, \mathbf{i} + \frac{1}{2}x \, \mathbf{j} \) on a particle moving once around the ellipse.

### 1. Area of the Ellipse
The area \( A \) of an ellipse is given by the formula:
\[
A = \pi \times a \times b
\]
where \( a \) and \( b \) are the semi-major and semi-minor axes of the ellipse, respectively.

### 2. Work Done by the Force Field
The work done \( W \) by a force field \( \mathbf{F}(x, y) \) on a particle moving along a curve \( C \) is given by:
\[
W = \oint_C \mathbf{F} \cdot d\mathbf{r}
\]
where \( d\mathbf{r} = (dx, dy) \) is the differential element of the curve.

For the given force field \( \mathbf{F}(x, y) = -\frac{1}{2}y \, \mathbf{i} + \frac{1}{2}x \, \mathbf{j} \) and a curve \( C \) described by the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the work done is:
\[
W = \oint_C \left(-\frac{1}{2}y \, dx + \frac{1}{2}x \, dy\right)
\]
We can rewrite this using Green's Theorem, which relates a line integral around a simple closed curve \( C \) to a double integral over the region \( R \) enclosed by \( C \):
\[
\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA
\]
where \( P = -\frac{1}{2}y \) and \( Q = \frac{1}{2}x \).

### 3. Application of Green's Theorem
The partial derivatives are:
\[
\frac{\partial Q}{\partial x} = \frac{1}{2}, \quad \frac{\partial P}{\partial y} = -\frac{1}{2}
\]
Thus,
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1
\]
The work done is then:
\[
W = \iint_R 1 \, dA
\]
This integral represents the area \( A \) of the ellipse \( R \), so:
\[
W = \text{Area of the ellipse} = \pi \times a \times b
\]
### Conclusion
The work done by the force field around the ellipse is equal to the area of the ellipse. Therefore, the area of the ellipse \( \pi ab \) is equal to the magnitude of the work done by the force field on a particle that moves once around the ellipse.

Would you like further details on any specific part of the solution?

### Related Questions:
1. How does Green's Theorem apply to other force fields?
2. What are the conditions under which Green's Theorem can be used?
3. How is the work done by a force field in three dimensions calculated?
4. Can the work done by the force field be negative? Why or why not?
5. What happens if the force field is conservative?
6. How does the force field influence the motion of a particle?
7. What are some physical interpretations of this problem in real-world scenarios?
8. How would the solution change if the ellipse were rotated?

### Tip:
Always remember to check the applicability of Green's Theorem by ensuring the vector field and the curve meet the necessary conditions!

Explore how the area of an ellipse relates to the work done by a specific force field. Learn about ellipse equations, line integrals, and Green's Theorem in this advanced mathematical analysis.

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