= Solution
On the specified parametrized <curve>, $\mathbf r(t)=(t^2,2t,t^3)$ and $\mathbf r'(t)=(2t,2,3t^2)$. Substitution into the field gives $\mathbf F(\mathbf r(t))=(4t^7,2t^8,8t^6)$. Hence
$$
\mathbf F\cdot\mathbf r'=8t^8+4t^8+24t^8=36t^8.
$$
The <line integral> is therefore
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r=\int_0^1 36t^8\,dt=4}.
$$
Both the field and the tangent vector must be evaluated along the <curve>; integrating one field component alone would miss two contributions.
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