Question

Find a vector field \(\mathbf{F}\) with the given curl. In each case, is the vector field you found unique? $$\operatorname{curl} \mathbf{F}=\langle 0,1,0\rangle$$

Short answer

Answer: A possible vector field \(\mathbf{F}\) for the given curl is given by: $$\mathbf{F} = \left\langle \frac{\partial g(x,z)}{\partial x}, 1, \frac{\partial g(x,z)}{\partial z} \right\rangle,$$ where \(g(x,z)\) is a function of \(x\) and \(z\). The vector field found is not unique, as the function \(g(x,z)\) can vary, leading to multiple different vector fields \(\mathbf{F}\) for the given curl.

Step-by-step solution

Define the given curl

We are given the curl of a vector field \(\mathbf{F}\): $$\operatorname{curl} \mathbf{F}=\langle 0,1,0\rangle$$

Integrate each component of the curl

In order to find a potential function \(\phi\) that is associated with the vector field \(\mathbf{F}\), we need to integrate each component of the given curl with respect to the corresponding variables: $$\phi_x = \int (\operatorname{curl}\mathbf{F})_i dx \implies \phi_x = \int 0 dx = 0$$ $$\phi_y = \int (\operatorname{curl}\mathbf{F})_j dy \implies \phi_y = \int 1 dy = y + g(x,z)$$ $$\phi_z = \int (\operatorname{curl}\mathbf{F})_k dz \implies \phi_z = \int 0 dz = 0$$

Combine components to find the potential function

We can now combine the components of \(\phi\) to find the associated potential function \(\phi(x,y,z)\): $$\phi(x,y,z) = y + g(x,z)$$

Compute the gradient of the potential function

In order to find the vector field \(\mathbf{F}\), we need to compute the gradient of the potential function \(\phi(x,y,z)\). This is given by: $$\mathbf{F} = \nabla \phi = \left\langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right\rangle$$ Now compute the gradients: $$\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x}(y + g(x,z)) = \frac{\partial g(x,z)}{\partial x}$$ $$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(y + g(x,z)) = 1$$ $$\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z}(y + g(x,z)) = \frac{\partial g(x,z)}{\partial z}$$

Finalize the vector field

By substituting the gradients of \(\phi\) back into the formula for the vector field \(\mathbf{F}\), we get: $$\mathbf{F} = \left\langle \frac{\partial g(x,z)}{\partial x}, 1, \frac{\partial g(x,z)}{\partial z} \right\rangle$$

Discuss the uniqueness of the found vector field

In our solution, the vector field \(\mathbf{F}\) depends on the function \(g(x,z)\). As the function \(g(x,z)\) can vary (even by constants), it means that there can be multiple different vector fields \(\mathbf{F}\) for the given curl. Therefore, the vector field we found is not unique.