The problem:
Compute the line integral $$\int_{\gamma} F \,\mathrm d r$$ where $F$ is a vector field $$F(x, y, z) = \left( y^2+z^4+x, 2xy-z^2+x, x^{18}y^6+z^{2016} \right)$$ and $\gamma$ is the bound of the intersection $T_1 \cap T_2$ where $$T_1 = \big\{(x, y, z) \in \mathbb{R}^3 \mid x^2+y^2\leq16 \big\}$$ and $$T_2 = \big\{(x, y, z)\in \mathbb{R}^3 \mid z=4, 0\leq y \leq 3 \big\}$$ The orientation of the curve is clockwise if looked at from the point $(0, 0, 100)$.
Since the given vector field seems highly complicated I immediately assumed that Stokes' theorem should be used. So the curve that we're integrating over is the circle $x^2+y^2\leq16$ on the height of $z=4$ where $0\leq y \leq 3$. If looked at from the top it looks something like this:
Now the surface that this encloses can be parametrized as: $$S: r(x, y)=(x, y, 4), \text{where } (x, y)\in D=\big\{(x, y)\in \mathbb{R^2}|x^2+y^2\leq16, 0 \leq y \leq 3 \big\}$$
The normal vector of this surface is $r'_x \times r'_y=(0, 0, 1)$. Since, after applying Stokes' theorem, I'm going to be doing a scalar product of $(\nabla \times F)(r(x, y)) \cdot (r'_x \times r'_y) = (\nabla \times F)(r(x, y)) \cdot (0, 0, 1)$, I've concluded that I don't care about the first two components of $\nabla \times F$, therefore I've computed it as: $$\nabla \times F=(P, Q, \frac{d}{dx}(2xy-z^2+x) - \frac{d}{dy}(y^2+z^4+x)) = (P, Q, 2y+1-2y) = (P, Q, 1) = G$$
By the way, the $P$ and $Q$ are not meant to represent the original components of the vector field $F$ but rather just placeholders.
And since Stokes' theorem says that the curve that encloses the surface should be oriented such that, when traversing the curve, the surface remains "on the left side", we'll add a minus sign, getting:
$$\int_{\gamma}Fdr = - \iint_S{(\nabla \times F)dS} = -\iint_S{GdS} = -\iint_D{G(r(x, y)) \cdot (0, 0, 1)dxdy} = -\iint_D{(P(r(x, y), Q(r(x, y), 1) \cdot (0, 0, 1)}dxdy = -\iint_Ddxdy = - A(D)$$
Computing the area of the region $D$ gives some really uncomfortable results. I've done it by subtracting the area of the little 'cap' on the top (the part that gets cut off by the line $y=3$) from the area of the half-circle. Furthermore I've gotten the area of the cap by subtracting the area of the triangle formed by points $(0, 0), (\sqrt{7}, 3), (-\sqrt{7}, 3)$ from the area of the "pizza-slice" (don't know the English word for it) enclosed by the circle and lines $y=\frac{\pm \sqrt{7}}{3}$. All that process got me to a really ugly solution: $$= 8\pi\bigg(1-\pi+2\arctan\bigg(\frac{3}{\sqrt7}\bigg)\bigg) + 3\sqrt7$$
So the final solution should be $-1$ times that.
This is the first problem in which I've actually applied Stokes' theorem and you can see why I'm being skeptical about whether I've done it correctly.
So, does my process seem correct? (also concerned about the orientation part aside from the very complicated result).
Thanks.
Your process is correct. As a future strategy probably the easiest way to do the area integral is by symmetry in polar coordinates. The integral becomes:
$$A(D) = 2\int_{\sin^{-1}\left(\frac{3}{4}\right)}^{\frac{\pi}{2}} \int_0^{3\csc \theta} rdrd\theta + 2\int_0^{\sin^{-1}\left(\frac{3}{4}\right)} \int_0^4 rdrd\theta$$
$$= \int_{\sin^{-1}\left(\frac{3}{4}\right)}^{\frac{\pi}{2}}9 \csc^2 \theta d\theta + 16\sin^{-1}\left(\frac{3}{4}\right) = 9\cot\left(\sin^{-1}\left(\frac{3}{4}\right)\right) + 16\sin^{-1}\left(\frac{3}{4}\right)$$
$$\implies A(D) = 3\sqrt{7} +16\sin^{-1}\left(\frac{3}{4}\right)$$