Calculate the work done by the force F = 3x2 i + (2x-y) j + z k along the curve parameterized by x = 2t2, y = t, z = 4t2 − t, from t=0 to t=1.

The work done $W$ by a force $\mathbf{F}$ along a curve $C$ is given by the line integral:

$$W = \int_C \mathbf{F} \cdot d\mathbf{r}$$

The force is given as $\mathbf{F} = 3x^2 \mathbf{i} + (2x-y) \mathbf{j} + z \mathbf{k}$.
The curve is parameterized by $x = 2t^2$, $y = t$, $z = 4t^2 - t$, from $t=0$ to $t=1$.

Substitute $x = 2t^2$, $y = t$, and $z = 4t^2 - t$ into the expression for $\mathbf{F}$:

$$F_x = 3x^2 = 3(2t^2)^2 = 3(4t^4) = 12t^4$$ $$F_y = 2x-y = 2(2t^2) - t = 4t^2 - t$$ $$F_z = z = 4t^2 - t$$

So, $\mathbf{F}(t) = 12t^4 \mathbf{i} + (4t^2 - t) \mathbf{j} + (4t^2 - t) \mathbf{k}$.

The position vector is $\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}$.

$$\mathbf{r}(t) = 2t^2 \mathbf{i} + t \mathbf{j} + (4t^2 - t) \mathbf{k}$$

Now, find the differential $d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt$:

$$\frac{dx}{dt} = \frac{d}{dt}(2t^2) = 4t$$ $$\frac{dy}{dt} = \frac{d}{dt}(t) = 1$$ $$\frac{dz}{dt} = \frac{d}{dt}(4t^2 - t) = 8t - 1$$

So, $d\mathbf{r} = (4t \mathbf{i} + 1 \mathbf{j} + (8t - 1) \mathbf{k}) dt$.

$$\mathbf{F} \cdot d\mathbf{r} = (F_x \frac{dx}{dt} + F_y \frac{dy}{dt} + F_z \frac{dz}{dt}) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = \left( (12t^4)(4t) + (4t^2 - t)(1) + (4t^2 - t)(8t - 1) \right) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = \left( 48t^5 + 4t^2 - t + (32t^3 - 4t^2 - 8t^2 + t) \right) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = \left( 48t^5 + 4t^2 - t + 32t^3 - 12t^2 + t \right) dt$$

Combine like terms:

$$\mathbf{F} \cdot d\mathbf{r} = \left( 48t^5 + 32t^3 + (4t^2 - 12t^2) + (-t + t) \right) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = (48t^5 + 32t^3 - 8t^2) dt$$

$$W = \int_0^1 (48t^5 + 32t^3 - 8t^2) dt$$ $$W = \left[ \frac{48t^6}{6} + \frac{32t^4}{4} - \frac{8t^3}{3} \right]_0^1$$ $$W = \left[ 8t^6 + 8t^4 - \frac{8}{3}t^3 \right]_0^1$$

$$W = \left( 8(1)^6 + 8(1)^4 - \frac{8}{3}(1)^3 \right) - \left( 8(0)^6 + 8(0)^4 - \frac{8}{3}(0)^3 \right)$$ $$W = \left( 8 + 8 - \frac{8}{3} \right) - (0)$$ $$W = 16 - \frac{8}{3}$$

To subtract, find a common denominator:

$$W = \frac{16 \times 3}{3} - \frac{8}{3}$$ $$W = \frac{48}{3} - \frac{8}{3}$$ $$W = \frac{40}{3}$$

The work done is $\frac{40}{3}$.