Consider the vector field:
$$\vec F = ye^x \hat i + (x^2 + e^x) \hat j + z^2e^z \hat k$$
A closed curve $C$ lies in the plane $x + y + z = 3$, oriented counterclockwise. The parametric representation of this curve is defined as:
$$C = (1+\ cost) \hat i + (1+\ sint) \hat j + (1 - \ cost - \ sint) \hat k$$
Where $t\in [0, 2\pi]$.
Let's check Stokes' Theorem:
$$\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \oint_{C} \vec F \cdot d \vec l$$
Let's start with the LHS:
$$\vec \nabla \times \vec F = 2x \hat k$$
The projection of the curve $C$ on the $xy$ plane is the circle with radius $1$ and center $(1, 1)$. Then:
$$d \vec a = \hat k dxdy$$
$$\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \int_{a} 2x dxdy = 2\int_{0}^{2\pi}\int_{0}^{1} r^2cos \theta drd \theta = 0$$
OK, so now we know:
$$\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \oint_{C} \vec F \cdot d \vec l = 0$$
It's time to check the RHS:
We gotta evaluate the following line integral:
$$\oint_{C} \vec F \cdot d \vec l = \oint_{C}[ye^x dx + (x^2 + e^x)dy + z^2e^z dz]$$
Where:
$$x = 1 + \ cost$$
$$y = 1 + \ sint$$
$$z = 1 - \ cost - \ sint$$
Then we get:
$$\oint_{C} \vec F \cdot d \vec l = C_1 + C_2 + C_3$$
Where:
$$C_1 = \int_{0}^{2\pi} (1+\ sint) e^{1+\ cost}(-\ sint)dt$$
$$C_2 = \int_{0}^{2\pi} [(1+\ cost)^2 + e^{1+\ cost}](\ cost)dt$$
$$C_3 = \int_{0}^{2\pi} (1-\ cost-\ sint)^2e^{1-\ cost-\ sint}(\ sint-\ cost)dt$$
But how can we evaluate these integrals?
I tried to use an Integral Solver but did not find primitives for them... If they have no exact solution, is there a numerical method to evaluate them? We at least know (if I did not get wrong evaluating the LHS) that:
$$\oint_{C} \vec F \cdot d \vec l = C_1 + C_2 + C_3 = 0$$