= Solution
For a smooth <vector field> near an oriented piecewise smooth surface $S$, <Stokes theorem> states
$$
\oint_{\partial S}\mathbf A\cdot d\mathbf x
=\int_S(\nabla\times\mathbf A)\cdot d\mathbf S,
$$
where the boundary orientation is induced by the surface <unit normal> through the right-hand rule. For a constant vector $\mathbf k$, the <curl> of $\phi\mathbf k$ is $\nabla\phi\times\mathbf k$. Apply <Stokes theorem> and use the scalar triple product:
$$
\mathbf k\cdot\oint_{\partial S}\phi\,d\mathbf x
=\int_S(\nabla\phi\times\mathbf k)\cdot d\mathbf S
=\mathbf k\cdot\int_Sd\mathbf S\times\nabla\phi.
$$
Since this holds for every constant $\mathbf k$, equality of all vector components proves the <vector-valued Stokes identity>
$$
\boxed{\int_Sd\mathbf S\times\nabla\phi=\oint_{\partial S}\phi\,d\mathbf x}.
$$
For the specified field, the two axis segments contribute zero because $x^2y^4$ vanishes on them. On the arc the direct <line integral> is
$$
I=\int_0^{\pi/2}\cos^2\theta\sin^4\theta(\cos\theta-\sin\theta)\,d\theta.
$$
Its two terms give
$$
\int_0^{\pi/2}\cos^3\theta\sin^4\theta\,d\theta
=\int_0^1u^4(1-u^2)\,du=\frac2{35},
$$
and
$$
\int_0^{\pi/2}\cos^2\theta\sin^5\theta\,d\theta
=\int_0^1u^2(1-u^2)^2\,du=\frac8{105}.
$$
Therefore
$$
\boxed{I=-\frac2{105}}.
$$
One can also check the orientation and sign by <Stokes theorem>. For the upward-oriented quarter disk, $(\nabla\times\mathbf A)_z=2xy^4-4x^2y^3$. Polar integration gives
$$
\int_0^1r^6dr\int_0^{\pi/2}(2\cos\theta\sin^4\theta-4\cos^2\theta\sin^3\theta)d\theta
=\frac17\left(\frac25-\frac8{15}\right)=-\frac2{105},
$$
consistent with the positively oriented boundary in the question.
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