91Ó°ÊÓ

The divergence of a vector field \( \mathbf{F}(x, y, z) = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).Given \( \mathbf{F}=2 x y e^{z} \mathbf{i}+e^{z} x^{2} \mathbf{j}+(x^{2} y e^{z}+z^{2}) \mathbf{k} \), we identify:- \( P = 2xy e^z \)- \( Q = e^z x^2 \)- \( R = x^2 y e^z + z^2 \)Now, calculate each term:1. \( \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(2xy e^z) = 2y e^z \)2. \( \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(e^z x^2) = 0 \)3. \( \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(x^2 y e^z + z^2) = x^2 y e^z + 2z \)Combine these to get the divergence:\( abla \cdot \mathbf{F} = 2y e^z + 0 + (x^2 y e^z + 2z) = 2y e^z + x^2 y e^z + 2z \).

The curl of a vector field \( \mathbf{F}(x, y, z) = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \]Calculate each component:1. \( \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(x^2 y e^z + z^2) = x^2 e^z \)2. \( \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(e^z x^2) = e^z x^2 \)3. \( \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(x^2 y e^z + z^2) = 2xy e^z \)4. \( \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(2xy e^z) = 2xy e^z \)5. \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^z x^2) = 2e^z x \)6. \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2xy e^z) = 2x e^z \)Simplify:- \( abla \times \mathbf{F} = (x^2e^z - e^z x^2) \mathbf{i} - (2xy e^z - 2xy e^z) \mathbf{j} + (2e^z x - 2xe^z) \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} \).Thus, \( abla \times \mathbf{F} = 0 \).

To find if \( \mathbf{F} \) is conservative, \( abla \times \mathbf{F} \) must be zero, which we have found in the previous step.To find \( f(x, y, z) \) such that \( abla f = \mathbf{F} \), we need to solve the following system:1. \( \frac{\partial f}{\partial x} = 2xy e^z \)2. \( \frac{\partial f}{\partial y} = e^z x^2 \)3. \( \frac{\partial f}{\partial z} = x^2 y e^z + z^2 \)Integrate the first equation with respect to \( x \):\( f(x, y, z) = \int 2xy e^z \, dx = x^2y e^z + g(y, z) \)Where \( g(y, z) \) is an arbitrary function.Now differentiate it with respect to \( y \) and match it with \( \frac{\partial f}{\partial y} = e^z x^2 \):\( \frac{\partial}{\partial y}(x^2y e^z + g(y, z)) = x^2 e^z + \frac{\partial g}{\partial y} = e^z x^2 \)Therefore, \( \frac{\partial g}{\partial y} = 0 \) implies \( g(y, z) = h(z) \) where \( h(z) \) is an arbitrary function.Now differentiate the expression \( f(x, y, z) = x^2y e^z + h(z) \) with respect to \( z \) and match it:\( x^2y e^z + \frac{dh}{dz} = x^2 y e^z + z^2 \)This means \( \frac{dh}{dz} = z^2 \) leading to \( h(z) = \frac{z^3}{3} + C \).Thus, \( f(x, y, z) = x^2y e^z + \frac{z^3}{3} + C \).