1. **State the problem:** We are given a vector field $$\mathbf{F}(x,y,z) = (5 \ln y - yz) \mathbf{i} + \left(\frac{5x}{y} - xz\right) \mathbf{j} + (-xy - \pi) \mathbf{k}$$. We need to show that $$\mathbf{F}$$ is conservative and find its potential function $$f(x,y,z)$$. 2. **Recall the definition:** A vector field $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$$ is conservative if there exists a scalar function $$f$$ such that $$\nabla f = \mathbf{F}$$. This means: $$\frac{\partial f}{\partial x} = P, \quad \frac{\partial f}{\partial y} = Q, \quad \frac{\partial f}{\partial z} = R.$$ Also, for $$\mathbf{F}$$ to be conservative in a simply connected domain, the curl must be zero: $$\nabla \times \mathbf{F} = \mathbf{0}.$$ 3. **Identify components:** $$P = 5 \ln y - yz,$$ $$Q = \frac{5x}{y} - xz,$$ $$R = -xy - \pi.$$ 4. **Check curl $$\nabla \times \mathbf{F}$$:** $$\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}.$$ Calculate each term: - $$\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(-xy - \pi) = -x,$$ - $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z} \left( \frac{5x}{y} - xz \right) = -x,$$ - $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(-xy - \pi) = -y,$$ - $$\frac{\partial P}{\partial z} = \frac{\partial}{\partial z} (5 \ln y - yz) = -y,$$ - $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( \frac{5x}{y} - xz \right) = \frac{5}{y} - z,$$ - $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (5 \ln y - yz) = \frac{5}{y} - z.$$ Substitute: $$\nabla \times \mathbf{F} = ( -x - (-x) ) \mathbf{i} - ( -y - (-y) ) \mathbf{j} + \left( \frac{5}{y} - z - \left( \frac{5}{y} - z \right) \right) \mathbf{k} = \mathbf{0}.$$ Since the curl is zero, $$\mathbf{F}$$ is conservative. 5. **Find potential function $$f$$:** We have: $$\frac{\partial f}{\partial x} = 5 \ln y - yz,$$ $$\frac{\partial f}{\partial y} = \frac{5x}{y} - xz,$$ $$\frac{\partial f}{\partial z} = -xy - \pi.$$ Integrate $$P$$ with respect to $$x$$: $$f(x,y,z) = \int (5 \ln y - yz) \, dx = (5 \ln y - yz) x + g(y,z),$$ where $$g(y,z)$$ is an arbitrary function of $$y,z$$. 6. **Differentiate $$f$$ with respect to $$y$$ and equate to $$Q$$:** $$\frac{\partial f}{\partial y} = x \frac{\partial}{\partial y} (5 \ln y - yz) + \frac{\partial g}{\partial y} = x \left( \frac{5}{y} - z \right) + g_y(y,z).$$ Set equal to $$Q$$: $$x \left( \frac{5}{y} - z \right) + g_y(y,z) = \frac{5x}{y} - xz,$$ which simplifies to $$g_y(y,z) = 0.$$ So $$g$$ does not depend on $$y$$, i.e., $$g = h(z)$$. 7. **Differentiate $$f$$ with respect to $$z$$ and equate to $$R$$:** $$\frac{\partial f}{\partial z} = x \frac{\partial}{\partial z} (5 \ln y - yz) + h'(z) = x(-y) + h'(z) = -xy + h'(z).$$ Set equal to $$R$$: $$-xy + h'(z) = -xy - \pi,$$ which gives $$h'(z) = -\pi.$$ 8. **Integrate $$h'(z)$$:** $$h(z) = -\pi z + C,$$ where $$C$$ is a constant. 9. **Write the potential function:** $$f(x,y,z) = x(5 \ln y - yz) - \pi z + C.$$ **Final answer:** $$\boxed{f(x,y,z) = x(5 \ln y - yz) - \pi z + C}.$$