1. **State the problem:** Given the vector field $$\mathbf{F} = \left( \frac{y}{\sqrt{1-x^2 y^2}} + 2xy^3 + 6 \right) \mathbf{i} + \left( \frac{x}{\sqrt{1-x^2 y^2}} + 3x^2 y^2 + 7 \right) \mathbf{j}$$ (a) Determine if $$\mathbf{F}$$ is conservative, and if yes, find the potential function $$\phi$$. (b) Find the work done by $$\mathbf{F}$$ when moving a particle from $$(-1,-1)$$ to $$(1,1)$$ along a straight line. 2. **Check if $$\mathbf{F}$$ is conservative:** For a vector field $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$$, if $$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$$ in a simply connected domain, then $$\mathbf{F}$$ is conservative. Given: $$P = \frac{y}{\sqrt{1-x^2 y^2}} + 2xy^3 + 6$$ $$Q = \frac{x}{\sqrt{1-x^2 y^2}} + 3x^2 y^2 + 7$$ Calculate $$\frac{\partial P}{\partial y}$$: First term: $$\frac{\partial}{\partial y} \left( \frac{y}{\sqrt{1-x^2 y^2}} \right)$$ Use quotient and chain rule: Let $$f(y) = y$$ and $$g(y) = \sqrt{1-x^2 y^2} = (1 - x^2 y^2)^{1/2}$$. $$\frac{d}{dy} \frac{f}{g} = \frac{g \cdot f' - f \cdot g'}{g^2}$$ Calculate derivatives: $$f' = 1$$ $$g' = \frac{1}{2} (1 - x^2 y^2)^{-1/2} \cdot (-2 x^2 y) = - \frac{x^2 y}{\sqrt{1-x^2 y^2}}$$ Therefore, $$\frac{\partial}{\partial y} \left( \frac{y}{\sqrt{1-x^2 y^2}} \right) = \frac{ \sqrt{1-x^2 y^2} \cdot 1 - y \left(- \frac{x^2 y}{\sqrt{1-x^2 y^2}} \right)}{1 - x^2 y^2} = \frac{\sqrt{1-x^2 y^2} + \frac{x^2 y^2}{\sqrt{1-x^2 y^2}} }{1 - x^2 y^2}$$ This simplifies to $$\frac{1}{(1-x^2 y^2)^{3/2}}$$ Next terms: $$\frac{\partial}{\partial y}(2 x y^{3}) = 6 x y^{2}$$ $$\frac{\partial}{\partial y}(6) = 0$$ So, $$\frac{\partial P}{\partial y} = \frac{1}{(1-x^2 y^2)^{3/2}} + 6 x y^{2}$$ Calculate $$\frac{\partial Q}{\partial x}$$: First term: $$\frac{\partial}{\partial x} \left( \frac{x}{\sqrt{1-x^2 y^2}} \right)$$ Similarly, Using quotient rule: $$f(x) = x, f' = 1$$ $$g(x) = \sqrt{1 - x^2 y^2}, g' = \frac{1}{2} (1 - x^2 y^2)^{-1/2} \cdot (-2x y^2) = - \frac{x y^2}{\sqrt{1-x^2 y^2}}$$ Therefore, $$\frac{\partial}{\partial x} \left( \frac{x}{\sqrt{1-x^2 y^2}} \right) = \frac{ \sqrt{1-x^2 y^2} \cdot 1 - x \left(- \frac{x y^2}{\sqrt{1-x^2 y^2}} \right)}{1-x^2 y^2} = \frac{\sqrt{1-x^2 y^2} + \frac{x^2 y^2}{\sqrt{1-x^2 y^2}}}{1-x^2 y^2} = \frac{1}{(1-x^2 y^2)^{3/2}}$$ Next terms: $$\frac{\partial}{\partial x}(3 x^{2} y^{2}) = 6 x y^{2}$$ $$\frac{\partial}{\partial x}(7) = 0$$ So, $$\frac{\partial Q}{\partial x} = \frac{1}{(1-x^2 y^2)^{3/2}} + 6 x y^{2}$$ 3. Since $$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$$, the field $$\mathbf{F}$$ is conservative. 4. **Find potential function $$\phi(x,y)$$ such that $$\nabla \phi = \mathbf{F}$$:** Integrate $$P$$ w.r.t. $$x$$: $$\phi(x,y) = \int P dx = \int \left( \frac{y}{\sqrt{1-x^{2} y^{2}}} + 2 x y^{3} + 6 \right) dx$$ Break integral: - $$\int \frac{y}{\sqrt{1-x^{2} y^{2}}} dx$$ - $$\int 2 x y^{3} dx = y^{3} x^{2} + C(y)$$ - $$\int 6 dx = 6x + C(y)$$ For the first integral, treat $$y$$ as constant: Substitute $$u = x y$$, then $$x = u/y$$ and $$dx = du / y$$. Then $$\int \frac{y}{\sqrt{1 - x^{2} y^{2}}} dx = \int \frac{y}{\sqrt{1 - u^{2}}} \frac{du}{y} = \int \frac{1}{\sqrt{1-u^{2}}} du = \arcsin u + C = \arcsin (x y)$$ 5. Summing all parts: $$\phi(x,y) = \arcsin(x y) + x^{2} y^{3} + 6 x + h(y)$$ 6. Find $$h(y)$$ by matching $$\frac{\partial \phi}{\partial y} = Q$$: Compute: $$\frac{\partial \phi}{\partial y} = \frac{1}{\sqrt{1 - x^{2} y^{2}}} \cdot x + 3 x^{2} y^{2} + h'(y)$$ Given $$Q = \frac{x}{\sqrt{1-x^{2} y^{2}}} + 3 x^{2} y^{2} + 7$$ Equate: $$\frac{\partial \phi}{\partial y} = Q \implies h'(y) = 7$$ Integrate: $$h(y) = 7 y + C$$ Thus, $$\phi(x,y) = \arcsin(x y) + x^{2} y^{3} + 6 x + 7 y + C$$ 7. **Calculate work done:** Work done by conservative field equals potential difference: $$W = \phi(1,1) - \phi(-1,-1)$$ Calculate: $$\phi(1,1) = \arcsin(1 \cdot 1) + 1^{2} \cdot 1^{3} + 6 (1) + 7 (1) + C = \frac{\pi}{2} + 1 + 6 + 7 + C = \frac{\pi}{2} + 14 + C$$ $$\phi(-1,-1) = \arcsin((-1)(-1)) + (-1)^{2} (-1)^{3} + 6 (-1) + 7 (-1) + C = \arcsin(1) + 1 \cdot (-1) - 6 -7 + C = \frac{\pi}{2} - 1 - 6 - 7 + C = \frac{\pi}{2} -14 + C$$ So, $$W = \left( \frac{\pi}{2} + 14 + C \right) - \left( \frac{\pi}{2} - 14 + C \right) = 14 + 14 = 28$$ **Final answers:** - Field $$\mathbf{F}$$ is conservative. - Potential function: $$\phi(x,y) = \arcsin(x y) + x^{2} y^{3} + 6 x + 7 y + C$$ - Work done from $$(-1,-1)$$ to $$(1,1)$$ is $$28$$.