The vector field \[ \mathrm{F}(x, y, z)=y z \mathbf{i}+x y^{2} \mathbf{j}+y z^{2} \mathrm{k} \] has divergence and curl

To find the divergence and curl of the vector field \(\mathrm{F}(x, y, z) = y z \mathbf{i} + x y^2 \mathbf{j} + y z^2 \mathbf{k}\), we will use the definitions of divergence and curl. ### Divergence The divergence of a vector field \(\mathrm{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}\) is given by: \[ \nabla \cdot \mathrm{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \] For our vector field, we identify: - \(P = y z\) - \(Q = x y^2\) - \(R = y z^2\) Now we compute each partial derivative: 1. \(\frac{\partial P}{\partial x} = \frac{\partial (y z)}{\partial x} = 0\) (since \(y\) and \(z\) are treated as constants with respect to \(x\)) 2. \(\frac{\partial Q}{\partial y} = \frac{\partial (x y^2)}{\partial y} = 2xy\) 3. \(\frac{\partial R}{\partial z} = \frac{\partial (y z^2)}{\partial z} = 2yz\) Now, we can sum these results to find the divergence: \[ \nabla \cdot \mathrm{F} = 0 + 2xy + 2yz = 2xy + 2yz \] ### Curl The curl of a vector field \(\mathrm{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}\) is given by: \[ \nabla \times \mathrm{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \] Now we compute each term: 1. \(\frac{\partial R}{\partial y} = \frac{\partial (y z^2)}{\partial y} = z^2\) 2. \(\frac{\partial Q}{\partial z} = \frac{\partial (x y^2)}{\partial z} = 0\) Thus, the \(i\)-component is: \[ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = z^2 - 0 = z^2 \] 3. \(\frac{\partial R}{\partial x} = \frac{\partial (y z^2)}{\partial x} = 0\) 4. \(\frac{\partial P}{\partial z} = \frac{\partial (y z)}{\partial z} = y\) Thus, the \(j\)-component is: \[ -\left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) = -\left( 0 - y \right) = y \] 5. \(\frac{\partial Q}{\partial x} = \frac{\partial (x y^2)}{\partial x} = y^2\) 6. \(\frac{\partial P}{\partial y} = \frac{\partial (y z)}{\partial y} = z\) Thus, the \(k\)-component is: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y^2 - z \] Putting it all together, we have: \[ \nabla \times \mathrm{F} = z^2 \mathbf{i} - y \mathbf{j} + (y^2 - z) \mathbf{k} \] ### Summary The divergence and curl of the vector field \(\mathrm{F}(x, y, z)\) are: \[ \nabla \cdot \mathrm{F} = 2xy + 2yz \] \[ \nabla \times \mathrm{F} = z^2 \mathbf{i} - y \mathbf{j} + (y^2 - z) \mathbf{k} \]