Let's solve the problem step by step: --- ## **Given:** \[ \vec{F}(x, y, z) = (3x^2y + z^3)\hat{i} + (x^3 - Ayz^2)\hat{j} + (-6y^2z + Bxz^2 + 2z)\hat{k} \] --- ### **(a) Find the values of the constants \( A \) and \( B \).** Since the field is **conservative**, \(\vec{F} = \nabla \phi\) So, the curl must be zero: \[ \nabla \times \vec{F} = 0 \] Let's compute the components: #### **1. \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \):** - \( F_1 = 3x^2y + z^3 \) - \( F_2 = x^3 - Ayz^2 \) \[ \frac{\partial F_2}{\partial x} = 3x^2 \] \[ \frac{\partial F_1}{\partial y} = 3x^2 \] \[ \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 3x^2 - 3x^2 = 0 \] #### **2. \( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \):** - \( F_3 = -6y^2z + Bxz^2 + 2z \) \[ \frac{\partial F_3}{\partial y} = -12yz \] \[ \frac{\partial F_2}{\partial z} = -A y \cdot 2z = -2A y z \] \[ \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = -12 y z - (-2A y z) = (-12 + 2A) y z \] Set this to zero for all \( y, z \): \[ -12 + 2A = 0 \implies 2A = 12 \implies \boxed{A = 6} \] #### **3. \( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \):** \[ \frac{\partial F_1}{\partial z} = 3z^2 \] \[ \frac{\partial F_3}{\partial x} = Bz^2 \] \[ \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = 3z^2 - Bz^2 = (3 - B)z^2 \] Set to zero for all \( z \): \[ 3 - B = 0 \implies \boxed{B = 3} \] --- ### **(b) Find a potential function \( \phi \) such that \( \vec{F} = \nabla \phi \), and \( \phi(1,1,1) = 5 \).** Now with \(A = 6\), \(B = 3\): \[ \vec{F} = (3x^2y + z^3)\hat{i} + (x^3 - 6y z^2)\hat{j} + (-6y^2z + 3x z^2 + 2z)\hat{k} \] We want to find \( \phi \) such that: \[ \frac{\partial \phi}{\partial x} = 3x^2y + z^3 \\ \frac{\partial \phi}{\partial y} = x^3 - 6y z^2 \\ \frac{\partial \phi}{\partial z} = -6y^2z + 3x z^2 + 2z \] #### **Step 1: Integrate w.r.t. \( x \):** \[ \phi(x, y, z) = \int (3x^2y + z^3) dx = x^3y + xz^3 + g(y, z) \] #### **Step 2: Differentiate w.r.t. \( y \):** \[ \frac{\partial \phi}{\partial y} = x^3 + \frac{\partial g}{\partial y} \] Set equal to given \( F_2 \): \[ x^3 + \frac{\partial g}{\partial y} = x^3 - 6y z^2 \implies \frac{\partial g}{\partial y} = -6y z^2 \] Integrate w.r.t. \( y \): \[ g(y, z) = -3y^2 z^2 + h(z) \] So, \[ \phi(x, y, z) = x^3y + xz^3 - 3y^2z^2 + h(z) \] #### **Step 3: Differentiate w.r.t. \( z \):** \[ \frac{\partial \phi}{\partial z} = 3xz^2 + x^3 \cdot 0 + x \cdot 3z^2 + (-3y^2)(2z) + h'(z) = 3x z^2 - 6y^2 z + h'(z) \] Set equal to \( F_3 \): \[ 3x z^2 - 6y^2 z + h'(z) = -6y^2z + 3x z^2 + 2z \] So, \[ h'(z) = 2z \implies h(z) = z^2 + C \] So the potential function is: \[ \phi(x, y, z) = x^3y + xz^3 - 3y^2z^2 + z^2 + C \] Now, use \( \phi(1, 1, 1) = 5 \): \[ \phi(1, 1, 1) = (1)(1) + (1)(1) - 3(1)(1) + 1 + C = 1 + 1 - 3 + 1 + C = 0 + C = 5 \implies C = 5 \] **Final potential function:** \[ \boxed{ \phi(x, y, z) = x^3y + xz^3 - 3y^2z^2 + z^2 + 5 } \] --- ### **(c) For the curve \( C \) parametrized by \( \vec{r}(t) = (5\cos t, 4\sin 3t, t) \), \( 0 < t < 2\pi \), compute the line integral \( \int_C \vec{F} \cdot d\vec{r} \).** For a conservative field, the line integral from point \( A \) to \( B \) is \( \phi(B) - \phi(A) \). Curve \( C \) goes from \( t=0 \) to \( t=2\pi \): \[ \vec{r}(0) = (5, 0, 0) \\ \vec{r}(2\pi) = (5, 0, 2\pi) \] So, \[ \int_C \vec{F} \cdot d\vec{r} = \phi(5, 0, 2\pi) - \phi(5, 0, 0) \] Calculate: #### **At \( (5, 0, 2\pi) \):** \[ \phi(5, 0, 2\pi) = (5)^3(0) + 5(2\pi)^3 - 3(0)^2(2\pi)^2 + (2\pi)^2 + 5 = 0 + 5 \cdot 8\pi^3 + 0 + 4\pi^2 + 5 = 40\pi^3 + 4\pi^2 + 5 \] #### **At \( (5, 0, 0) \):** \[ \phi(5, 0, 0) = (5)^3(0) + 5(0)^3 - 3(0)^2(0)^2 + 0^2 + 5 = 0 + 0 + 0 + 0 + 5 = 5 \] #### **Integral:** \[ \int_C \vec{F} \cdot d\vec{r} = (40\pi^3 + 4\pi^2 + 5) - 5 = \boxed{40\pi^3 + 4\pi^2} \] --- ## **Final Answers** --- **(a)** \( \boxed{A = 6, \quad B = 3} \) **(b)** \( \boxed{\phi(x, y, z) = x^3y + xz^3 - 3y^2z^2 + z^2 + 5} \) **(c)** \( \boxed{\int_C \vec{F} \cdot d\vec{r} = 40\pi^3 + 4\pi^2} \)