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We are asked to evaluate the integral \(\int_C \vec{F} \times d\vec{l}\), where the vector field is given as:

\[
\vec{F}(x, y, z) = xyi - zj + x^2k
\]

and the curve \(C\) is parameterized as:

\[
\begin{cases}
x = t^2 \\
y = 2t \\
z = t^3
\end{cases}
\quad \text{for} \quad t \in [0, 1].
\]

### Step 1: Compute the differential \(d\vec{l}\)
The position vector \(\vec{r}(t)\) is:

\[
\vec{r}(t) = x(t)i + y(t)j + z(t)k = t^2i + 2tj + t^3k.
\]

The differential \(d\vec{l} = \frac{d\vec{r}}{dt} dt\) is:

\[
d\vec{l} = \frac{d}{dt}(t^2 i + 2t j + t^3 k) dt = (2t i + 2 j + 3t^2 k) dt.
\]

### Step 2: Parametrize the vector field \(\vec{F}(x(t), y(t), z(t))\)
Substitute the parameterization of the curve into \(\vec{F}(x, y, z)\):

\[
\vec{F}(t) = (t^2)(2t) i - (t^3) j + (t^2)^2 k = 2t^3 i - t^3 j + t^4 k.
\]

### Step 3: Compute the cross product \(\vec{F}(t) \times d\vec{l}\)
Now compute the cross product \(\vec{F}(t) \times d\vec{l}\):

\[
\vec{F}(t) = 2t^3 i - t^3 j + t^4 k,
\]
and
\[
d\vec{l} = 2t i + 2 j + 3t^2 k.
\]

The cross product is:

\[
\vec{F}(t) \times d\vec{l} = \begin{vmatrix} 
i & j & k \\ 
2t^3 & -t^3 & t^4 \\ 
2t & 2 & 3t^2 
\end{vmatrix}.
\]

Expanding this determinant:

\[
\vec{F}(t) \times d\vec{l} = i \begin{vmatrix} -t^3 & t^4 \\ 2 & 3t^2 \end{vmatrix} - j \begin{vmatrix} 2t^3 & t^4 \\ 2t & 3t^2 \end{vmatrix} + k \begin{vmatrix} 2t^3 & -t^3 \\ 2t & 2 \end{vmatrix}.
\]

These 2x2 determinants simplify as follows:

- For the \(i\)-component:

\[
\begin{vmatrix} -t^3 & t^4 \\ 2 & 3t^2 \end{vmatrix} = (-t^3)(3t^2) - (t^4)(2) = -3t^5 - 2t^4 = -t^4(3t + 2).
\]

- For the \(j\)-component:

\[
\begin{vmatrix} 2t^3 & t^4 \\ 2t & 3t^2 \end{vmatrix} = (2t^3)(3t^2) - (t^4)(2t) = 6t^5 - 2t^5 = 4t^5.
\]

- For the \(k\)-component:

\[
\begin{vmatrix} 2t^3 & -t^3 \\ 2t & 2 \end{vmatrix} = (2t^3)(2) - (-t^3)(2t) = 4t^3 + 2t^4 = 2t^3(2 + t).
\]

Thus, the cross product is:

\[
\vec{F}(t) \times d\vec{l} = -t^4(3t + 2)i - 4t^5 j + 2t^3(2 + t)k.
\]

### Step 4: Integrate over \(t \in [0, 1]\)
Now we integrate each component from \(t = 0\) to \(t = 1\):

1. For the \(i\)-component:

\[
\int_0^1 -t^4(3t + 2) dt = -\int_0^1 (3t^5 + 2t^4) dt = -\left[\frac{3t^6}{6} + \frac{2t^5}{5}\right]_0^1 = -\left(\frac{1}{2} + \frac{2}{5}\right) = -\frac{9}{10}.
\]

2. For the \(j\)-component:

\[
\int_0^1 -4t^5 dt = -\left[\frac{4t^6}{6}\right]_0^1 = -\frac{2}{3}.
\]

3. For the \(k\)-component:

\[
\int_0^1 2t^3(2 + t) dt = \int_0^1 (4t^3 + 2t^4) dt = \left[\frac{4t^4}{4} + \frac{2t^5}{5}\right]_0^1 = 1 + \frac{2}{5} = \frac{7}{5}.
\]

### Final Answer:
The final result is:

\[
\int_C \vec{F} \times d\vec{l} = -\frac{9}{10}i - \frac{2}{3}j + \frac{7}{5}k.
\]

Thus, the correct option is:

\[
\boxed{d}.
\]

Would you like further details on any step? Here are 5 related questions:

1. How is the cross product of two vectors calculated?
2. Why is parameterization of the curve important in line integrals?
3. How do you compute a line integral of a vector field?
4. What are some physical interpretations of the cross product in vector calculus?
5. How can this problem be extended to a surface integral?

**Tip:** When working with vector fields and line integrals, always ensure correct parameterization and check the limits of integration carefully!

Evaluar la integral ∫_C (F × dl) dada el campo vectorial F(x, y, z) = xyi - zj + x²k y la curva C: x = t², y = 2t, z = t³, desde t = 0 hasta t = 1.

Learn how to evaluate the line integral ∫_C (F × dl) for the vector field F(x, y, z) = xyi - zj + x²k. The curve is parameterized as x = t², y = 2t, z = t³ from t = 0 to t = 1. This problem involves key vector calculus concepts like cross products and line integrals.

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