A vector field $\vec F$ is defined on the disk $D$ of radius $5$
$$\vec F=\big[-y^3+y\sin(xy)\big]\hat{x}+\big[4x(1-y^2)+x\sin(xy)\big]\hat{y}$$
Consider the line integral
$$ \int_C \vec F\,\mathrm{d}r $$
where $C$ is some closed curve contained in $D$. For which $C$ is this integral the largest?
We used Green's Theorem and found that $-2≤y≤2$, so $C$ is the boundary of the region bounded by $y=2$, $y=-2$ $\cup$ two circular arcs, and this maximizes the line integral. I don't see it.
The question should have provided with the orientation. In fact, the line integral is largest for a clockwise oriented curve.
If $\vec F = (P, Q)$. For a clockwise oriented curve and given vector field, $P_y - Q_x = y^2-4$. We note that $(y^2-4)$ is negative for $|y| \lt 2$ and positive for $ |y| \gt 2$. I will use it later in the answer.
If we take a closed curve which is boundary of the disk, it is easy to see that,
$\displaystyle \int_0^{2\pi} \int_0^5 r (r^2\sin^2\theta - 4) \ dr \ d\theta = \frac{225\pi}{4} \approx 176.715$
[Now we could find a piece-wise smooth simple closed curve oriented clockwise which avoids most of the region between $-2 \leq y \leq 2$ and that would make the integral even larger.]
If we put a restriction that it is a counterclockwise oriented closed curve as the book says in the hint, the integral becomes
$\displaystyle \int_C \vec F \cdot d\vec r = \iint_R (4-y^2) \ dA$
The below diagram depicts the closed curve that would maximize the line integral.
We know that $(4-y^2)$ is negative for $|y| \gt 2$. The closed curve depicted in the diagram avoids any point where $|y| \gt 2$ but also exploits max range of $-5 \leq x \leq 5$.
You can do the integral in couple different ways. I will do it over the rectangle in the first quadrant and then over region between line segment $AB$ and arc $AB$ above x-axis and then multiply the integral by $4$ due to symmetry.
$\displaystyle 4 \int_0^{\sqrt {21}} \int_0^2 (4 - y^2) \ dy \ dx + 4 \int_{\sqrt{21}}^5 \int_0^{\sqrt{25-x^2}} (4 - y^2) \ dy \ dx \approx 104.9293$