Let's analyze each part of the problem step-by-step:

(i) Explain why $\vec{F}$ is conservative on the domain $\mathcal{D} = \mathbb{R}^3$.

A vector field $\vec{F}$ is conservative if there exists a scalar potential function $g(x, y, z)$ such that $\vec{F} = \nabla g$. One equivalent condition for a vector field to be conservative in $\mathbb{R}^3$ is that its curl must be zero everywhere in the domain. So we check if $\nabla \times \vec{F} = \vec{0}$.

Given: $\vec{F} = \langle 3x^2 + 2y^2 + 3z, 4xy + 2y - z, 3x - y - z \rangle$

To compute $\nabla \times \vec{F}$, we use the definition of the curl: $\nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$ where $\vec{F} = \langle F_x, F_y, F_z \rangle$.

Calculating Components of the Curl

1. $\frac{\partial F_z}{\partial y} = -1$
2. $\frac{\partial F_y}{\partial z} = -1$

So, the first component of $\nabla \times \vec{F}$ is: $\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = -1 - (-1) = 0$

3. $\frac{\partial F_x}{\partial z} = 3$
4. $\frac{\partial F_z}{\partial x} = 3$

So, the second component of $\nabla \times \vec{F}$ is: $\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 3 - 3 = 0$

5. $\frac{\partial F_y}{\partial x} = 4y$
6. $\frac{\partial F_x}{\partial y} = 4y$

So, the third component of $\nabla \times \vec{F}$ is: $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 4y - 4y = 0$

Since all components of $\nabla \times \vec{F}$ are zero, we conclude: $\nabla \times \vec{F} = \vec{0}$ Therefore, $\vec{F}$ is conservative on $\mathbb{R}^3$.

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(ii) Find a potential function $g(x, y, z)$ for $\vec{F}$.

Since $\vec{F}$ is conservative, there exists a potential function $g(x, y, z)$ such that: $\vec{F} = \nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right)$ We are given: $F_x = \frac{\partial g}{\partial x} = 3x^2 + 2y^2 + 3z$ $F_y = \frac{\partial g}{\partial y} = 4xy + 2y - z$ $F_z = \frac{\partial g}{\partial z} = 3x - y - z$

To find $g(x, y, z)$, we integrate each component.

Integrating with respect to $x$:

Integrate $F_x$ with respect to $x$: $g(x, y, z) = \int (3x^2 + 2y^2 + 3z) \, dx = x^3 + 2xy^2 + 3xz + h(y, z)$ where $h(y, z)$ is a function of $y$ and $z$ alone.

Differentiating with respect to $y$:

$\frac{\partial g}{\partial y} = 4xy + \frac{\partial h}{\partial y}$ Comparing this with $F_y = 4xy + 2y - z$, we find: $\frac{\partial h}{\partial y} = 2y - z$ Integrating with respect to $y$: $h(y, z) = y^2 - yz + k(z)$ where $k(z)$ is a function of $z$ alone.

Differentiating with respect to $z$:

[ \frac{\partial g}{\partial z} = 3x - y - \frac{d}{dz}(yz) - \frac{d}{dz}