Answer :
Answer:
To determine if the vector field \(\vec{A} = (2x - yx) \mathbf{i} + (2y - xx) \mathbf{j} + (2x - xy) \mathbf{k}\) is irrotational (also known as conservative), we need to check if it satisfies the condition of being a conservative vector field, which means its curl is zero.
The curl of a vector field \(\vec{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}\) is given by:
\[ \nabla \times \vec{A} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) \mathbf{i} - \left( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \right) \mathbf{j} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \mathbf{k} \]
### Calculating the Curl:
Given:
\[ A_x = 2x - yx \]
\[ A_y = 2y - xx \]
\[ A_z = 2x - xy \]
Let's calculate the partial derivatives:
\[ \frac{\partial A_z}{\partial y} = \frac{\partial (2x - xy)}{\partial y} = 2 - x \]
\[ \frac{\partial A_y}{\partial z} = \frac{\partial (2y - xx)}{\partial x} = -2x \]
\[ \frac{\partial A_z}{\partial x} = \frac{\partial (2x - xy)}{\partial x} = 2 - y \]
\[ \frac{\partial A_x}{\partial z} = \frac{\partial (2x - yx)}{\partial y} = 0 \]
\[ \frac{\partial A_y}{\partial x} = \frac{\partial (2y - xx)}{\partial x} = -2x \]
\[ \frac{\partial A_x}{\partial y} = \frac{\partial (2x - yx)}{\partial y} = 2 - x \]
Now, let's calculate the curl:
\[ \nabla \times \vec{A} = \left( 2 - x - (-2x) \right) \mathbf{i} - \left( (2 - y) - 0 \right) \mathbf{j} + \left( (-2x) - (2 - x) \right) \mathbf{k} \]
\[ \nabla \times \vec{A} = (2 - x + 2x) \mathbf{i} - (2 - y) \mathbf{j} - (2x - 2 + x) \mathbf{k} \]
\[ \nabla \times \vec{A} = (2 + x) \mathbf{i} - (2 - y) \mathbf{j} - (x - 2) \mathbf{k} \]
### Checking if the Curl is Zero:
For the vector field to be irrotational, the curl (\( \nabla \times \vec{A} \)) must be zero.
\[ \nabla \times \vec{A} = (2 + x) \mathbf{i} - (2 - y) \mathbf{j} - (x - 2) \mathbf{k} \]
The curl is not zero since it has components dependent on \( x \) and \( y \).
So, the vector field \(\vec{A}\) is **not irrotational**.
The correct answer is **No**.