The problem in the image involves calculating the line integral of a conservative vector field $\vec{F}$ around a triangular path in the $xy$-plane. The vertices of the triangle are:

• $(0, 0)$ at the origin,
• $(a, 0)$ along the $x$-axis,
• $\left( \frac{a}{2}, a \right)$ along the diagonal pointing upwards.

The integral is expressed as:

$\oint_C \vec{F} \cdot d\vec{l}$

where $\oint_C$ represents a closed loop integral, meaning the path returns to the starting point.

Solution Outline:

1. Conservative Vector Field Property: For a conservative force field, the line integral over a closed loop is always zero. This is because the potential difference between the start and end point is zero, as the starting and ending points are the same.

   Therefore, we conclude:

   $\oint_C \vec{F} \cdot d\vec{l} = 0$

This result stems from the fundamental theorem of line integrals, which states that the integral of a conservative vector field over any closed path is zero.

Let me know if you'd like more details on the vector field or any other clarifications!

Additional Questions:

1. What is a conservative vector field, and how does it relate to potential energy?
2. How is the fundamental theorem of line integrals applied in vector calculus?
3. Can you explain how to parametrize each segment of a triangular path for a line integral?
4. How does the concept of curl relate to conservative fields?
5. Could you show an example where the line integral of a non-conservative vector field is non-zero?

Tip:

In a conservative field, the key to solving line integrals over closed paths is understanding that the field has zero curl and can be described as the gradient of a scalar potential function.