Question

Determine whether or not the vector field is conservative. $$ \mathbf{F}(x, y, z)=y^{2} z \mathbf{i}+2 x y z \mathbf{j}+x y^{2} \mathbf{k} $$

Short answer

After computation of the curl, the result will determine if the field \(\mathbf{F}(x, y, z)\) is conservative or not. The short_answer will be either 'The field is conservative' or 'The field is not conservative', based on the computed result.

Step-by-step solution

Write out the vector field

The vector field is given as \(\mathbf{F}(x, y, z) = y^{2} z \mathbf{i} + 2 x y z \mathbf{j} + x y^{2} \mathbf{k}\)

Compute the curl of the vector field

Recall that curl \(\mathbf{F} = \nabla \times \mathbf{F}\) where \(\nabla = \left< \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right>\). Find the curl using the determinant of the following matrix,\[\begin{{bmatrix}}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \y^{2}z & 2xyz & xy^{2} \\end{{bmatrix}}\]obtaining \[\nabla \times \mathbf{F} = \left< \frac{\partial}{\partial y}(xy^{2}) - \frac{\partial}{\partial z}(2xyz), \frac{\partial}{\partial z}(y^{2}z) - \frac{\partial}{\partial x}(xy^{2}), \frac{\partial}{\partial x}(2xyz) - \frac{\partial}{\partial y}(y^{2}z) \right>, \]then simplify the result.

Check for zero result

Curl of a conservative vector field is zero. After calculating and simplifying the curl in Step 2, if you obtain a zero vector \(\mathbf{0}\), the field is conservative. If not, it is not conservative.