Check whether the line integral of integral ++(1,2,3)(2,0,0)(xdx ydy zdz) is independent of path or not

To determine whether the line integral of the vector field given by the expression \( \mathbf{F} = (x, y, z) \) is independent of the path, we need to check if this vector field is conservative. A vector field is conservative if it can be expressed as the gradient of a scalar potential function, which implies that the line integral between two points is path-independent. To establish whether \( \mathbf{F} \) is conservative, we can use the following steps:

Step 1: Check the Conditions for Conservativeness

A vector field \( \mathbf{F} = (P, Q, R) \) is conservative if the following conditions are met:

• The field is defined in a simply connected domain.
• The curl of the vector field is zero: \( \nabla \times \mathbf{F} = \mathbf{0} \).

Step 2: Calculate the Curl of the Vector Field

For our vector field \( \mathbf{F} = (x, y, z) \), we identify:

• \( P = x \)
• \( Q = y \)
• \( R = z \)

The curl of \( \mathbf{F} \) is given by:

\nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x & y & z
\end{vmatrix}

Calculating this determinant, we find:

\nabla \times \mathbf{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}, \frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}, \frac{\partial y}{\partial x} - \frac{\partial x}{\partial y} \right) = (0 - 0, 0 - 0, 0 - 0) = (0, 0, 0).

Step 3: Analyze the Result

Since \( \nabla \times \mathbf{F} = (0, 0, 0) \), the curl is indeed zero. This indicates that the vector field is conservative. Additionally, if the domain is simply connected (which it is in this case, as we are dealing with all of \( \mathbb{R}^3 \)), we can conclude that the line integral of \( \mathbf{F} \) is independent of the path taken between two points.

Step 4: Conclusion

Thus, the line integral of the vector field \( (x, y, z) \) is independent of the path. This means that regardless of the route taken from point \( (1, 2, 3) \) to point \( (2, 0, 0) \), the value of the integral will remain the same, depending only on the endpoints. This property is particularly useful in physics and engineering, where it simplifies the calculation of work done by a force field.