Given the vector field $\vec{F} = \langle 3x^2 + 2y^2 + 3z, 4xy + 2y - z, 3x - y - z \rangle$, let's go through 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 it can be expressed as the gradient of a scalar potential function $g(x, y, z)$. A key indicator that a vector field is conservative is that the curl of $\vec{F}$ is zero. Thus, we need to verify if $\nabla \times \vec{F} = \vec{0}$.

The components of $\vec{F}$ are: $F_x = 3x^2 + 2y^2 + 3z, \quad F_y = 4xy + 2y - z, \quad F_z = 3x - y - z.$

We calculate the curl of $\vec{F}$: $\nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \hat{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \hat{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \hat{k}.$

After evaluating each component, if the result is zero, it confirms that $\vec{F}$ is conservative on $\mathbb{R}^3$.

(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 $\vec{F} = \nabla g$. To find $g(x, y, z)$, we integrate each component of $\vec{F}$ with respect to its corresponding variable, checking for consistency across partial derivatives.

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

2. Differentiate $g(x, y, z)$ with respect to $y$ and equate it to $F_y$: By continuing with this process, we determine the full form of $g(x, y, z)$ and verify it.

(iii) Compute the work done by $\vec{F}$ to move a particle from $(-1, 0, 1)$ to $(0, 0, 1)$.

For a conservative field, the work done by $\vec{F}$ along any path from point $A$ to point $B$ is given by the difference in potential between those points: $W = g(0, 0, 1) - g(-1, 0, 1).$

We substitute these points into the potential function $g(x, y, z)$ once it is found.

(iv) Does this work depend on the path used to get from $(-1, 0, 1)$ to $(0, 0, 1)$?

In a conservative vector field, the work done only depends on the endpoints of the path, not on the specific path taken. Therefore, the work done by $\vec{F}$ does not depend on the path from $(-1, 0, 1)$ to $(0, 0, 1)$.

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Would you like further details on any specific step? Here are some related questions:

1. How can one verify if a vector field is conservative in higher dimensions?
2. What are the practical applications of calculating work done in conservative fields?
3. How does one interpret potential functions in physics and engineering?
4. Why does the work in conservative fields depend only on endpoints?
5. Could we apply similar methods to compute work in non-conservative fields?

Tip: When dealing with conservative vector fields, always check if the curl is zero, as this simplifies finding a potential function immensely.