Divergence Lagrange Laplacian B0203A

1. **Problem Statement:** (a)(i) State the Divergence theorem without proof. (a)(ii) Calculate the outward flux of \( F(x,y,z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \) across the boundary \( S \) of the region cut from the solid cylinder \( x^2 + y^2 \leq a^2 \) by the planes \( z=0 \) and \( z=1 \). (b) Use Lagrange multipliers to find the maximum and minimum values of \( f(x,y,z) = x + 2y \) subject to constraints \( x + y + z = 1 \) and \( y^2 + z^2 = 4 \). (c) Find the Laplacian \( \nabla^2 F \) of \( F(x,y,z) = e^x \sin y \mathbf{i} + e^x \cos y \mathbf{j} + z \mathbf{k} \). --- 2. **(a)(i) Divergence Theorem Statement:** The Divergence theorem states: $$ \int\int_S \mathbf{F} \cdot d\mathbf{S} = \int\int\int_V (\nabla \cdot \mathbf{F}) \, dV $$ where \( S \) is the closed surface bounding volume \( V \), \( \mathbf{F} \) is a vector field, and \( \nabla \cdot \mathbf{F} \) is the divergence of \( \mathbf{F} \). 3. **(a)(ii) Calculate Outward Flux:** - Given \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \). - Compute divergence: $$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (z^2) = 2x + 2y + 2z $$ - The volume \( V \) is the cylinder \( x^2 + y^2 \leq a^2 \) between \( z=0 \) and \( z=1 \). - Set up the triple integral for flux: $$ \Phi = \int_0^1 \int_{x^2 + y^2 \leq a^2} (2x + 2y + 2z) \, dA \, dz $$ - By symmetry, \( \int_{x^2 + y^2 \leq a^2} x \, dA = 0 \) and \( \int_{x^2 + y^2 \leq a^2} y \, dA = 0 \). - So only \( 2z \) term contributes: $$ \Phi = \int_0^1 2z \left( \int_{x^2 + y^2 \leq a^2} dA \right) dz $$ - Area of circle \( x^2 + y^2 \leq a^2 \) is \( \pi a^2 \). - Thus: $$ \Phi = \int_0^1 2z \pi a^2 \, dz = 2 \pi a^2 \int_0^1 z \, dz = 2 \pi a^2 \left[ \frac{z^2}{2} \right]_0^1 = \pi a^2 $$ 4. **(b) Lagrange Multipliers:** - Objective function: \( f(x,y,z) = x + 2y \). - Constraints: \( g(x,y,z) = x + y + z - 1 = 0 \) \( h(x,y,z) = y^2 + z^2 - 4 = 0 \) - Set up Lagrangian: $$ L = x + 2y - \lambda (x + y + z - 1) - \mu (y^2 + z^2 - 4) $$ - Compute partial derivatives and set to zero: $$ \frac{\partial L}{\partial x} = 1 - \lambda = 0 \Rightarrow \lambda = 1 $$ $$ \frac{\partial L}{\partial y} = 2 - \lambda - 2\mu y = 0 \Rightarrow 2 - 1 - 2\mu y = 0 \Rightarrow 1 = 2\mu y $$ $$ \frac{\partial L}{\partial z} = -\lambda - 2\mu z = 0 \Rightarrow -1 - 2\mu z = 0 \Rightarrow 2\mu z = -1 $$ - From above: $$ 2\mu y = 1, \quad 2\mu z = -1 $$ - Divide equations: $$ \frac{y}{z} = \frac{1}{-1} = -1 \Rightarrow y = -z $$ - Use constraint \( y^2 + z^2 = 4 \): $$ y^2 + z^2 = z^2 + z^2 = 2z^2 = 4 \Rightarrow z^2 = 2 \Rightarrow z = \pm \sqrt{2} $$ - Then \( y = -z = \mp \sqrt{2} \). - Use constraint \( x + y + z = 1 \): $$ x = 1 - y - z = 1 - (-z) - z = 1 + z - z = 1 $$ - So points: $$ (x,y,z) = (1, -\sqrt{2}, \sqrt{2}) \quad \text{and} \quad (1, \sqrt{2}, -\sqrt{2}) $$ - Evaluate \( f \) at these points: $$ f(1, -\sqrt{2}, \sqrt{2}) = 1 + 2(-\sqrt{2}) = 1 - 2\sqrt{2} $$ $$ f(1, \sqrt{2}, -\sqrt{2}) = 1 + 2\sqrt{2} $$ - Maximum is \( 1 + 2\sqrt{2} \), minimum is \( 1 - 2\sqrt{2} \). 5. **(c) Find \( \nabla^2 F \):** - Given: $$ F = e^x \sin y \mathbf{i} + e^x \cos y \mathbf{j} + z \mathbf{k} $$ - The Laplacian operator applied component-wise: $$ \nabla^2 F = (\nabla^2 F_x) \mathbf{i} + (\nabla^2 F_y) \mathbf{j} + (\nabla^2 F_z) \mathbf{k} $$ - Compute \( \nabla^2 F_x = \nabla^2 (e^x \sin y) \): $$ \frac{\partial^2}{\partial x^2} (e^x \sin y) = e^x \sin y $$ $$ \frac{\partial^2}{\partial y^2} (e^x \sin y) = e^x \frac{\partial^2}{\partial y^2} (\sin y) = e^x (-\sin y) = -e^x \sin y $$ $$ \frac{\partial^2}{\partial z^2} (e^x \sin y) = 0 $$ - Sum: $$ \nabla^2 F_x = e^x \sin y - e^x \sin y + 0 = 0 $$ - Compute \( \nabla^2 F_y = \nabla^2 (e^x \cos y) \): $$ \frac{\partial^2}{\partial x^2} (e^x \cos y) = e^x \cos y $$ $$ \frac{\partial^2}{\partial y^2} (e^x \cos y) = e^x \frac{\partial^2}{\partial y^2} (\cos y) = e^x (-\cos y) = -e^x \cos y $$ $$ \frac{\partial^2}{\partial z^2} (e^x \cos y) = 0 $$ - Sum: $$ \nabla^2 F_y = e^x \cos y - e^x \cos y + 0 = 0 $$ - Compute \( \nabla^2 F_z = \nabla^2 (z) \): $$ \frac{\partial^2}{\partial x^2} (z) = 0, \quad \frac{\partial^2}{\partial y^2} (z) = 0, \quad \frac{\partial^2}{\partial z^2} (z) = 0 $$ - Sum: $$ \nabla^2 F_z = 0 $$ - Therefore: $$ \nabla^2 F = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} = \mathbf{0} $$ **Final answers:** - (a)(ii) Outward flux \( = \pi a^2 \). - (b) Maximum value \( = 1 + 2\sqrt{2} \), minimum value \( = 1 - 2\sqrt{2} \). - (c) \( \nabla^2 F = \mathbf{0} \).