By Green's theorem, the line integral of F along C is equal to the integral of the curl of F (two-dimensional curl, that is) over the region bounded by C, where C is a generic path that is oriented counterclockwise. However, our C run clockwise, so we multiply the following by -1.
[tex]\displaystyle \int_C \vec F \cdot d\vec r = -\iint_R \frac{\partial(4x^2+\sqrt y)}{\partial x} - \frac{\partial(\sqrt x+4y^3)}{\partial y} \, dx \, dy[/tex]
where R is the set
[tex]R = \left\{(x, y) : 0 \le x \le \pi \text{ and } 0 \le y \le \sin(x)\right\}[/tex]
Compute the double integral:
[tex]\displaystyle -\iint_R \frac{\partial(4x^2+\sqrt y)}{\partial x} - \frac{\partial(\sqrt x+4y^3)}{\partial y} \, dx \, dy = -\int_0^\pi \int_0^{\sin(x)} (8x - 12y^2) \, dy \, dx[/tex]
Integrating with respect to y is trivial:
[tex]\displaystyle \cdots = - \int_0^\pi (8x\sin(x) - 4\sin^3(x)) \, dx[/tex]
Integrating by parts with
u = x ⇒ du = dx
dv = sin(x) dx ⇒ v = -cos(x)
gives
[tex]\displaystyle \int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx = -x\cos(x) + \sin(x)[/tex]
while in the other integral, we have by substitution
[tex]\displaystyle \int \sin^3(x) \, dx = \int \sin(x) (1-\cos^2(x)) \, dx \\= - \int (1-\cos^2(x)) d(\cos(x)) \\ = -\cos(x) + \frac13 \cos^3(x)[/tex]
Then our last integral evaluates to
[tex]\displaystyle \cdots = - \left(8\pi - 4 + \frac43\right) + \left(4 - \frac43\right) = \boxed{\frac{16}3-8\pi}[/tex]
