$\int_{K}\left(x^{2}+y z\right) d x+\left(y^{2}+x z\right) d y+\left(z^{2}-x y\right) d z$ where K is a closed curve, oriented positive, consisting of an arc defined by the parametric equation $x=a\cos t, y=a\sin t, z=\frac{1}{2\pi}t$ and segment $BA,A=(a,0,0),B=(a,0,1)$
I drew this K-curve for the parameter a = 2 to see what area we would integrate over.
That is, looking from above, it will be a circle of radius: $$ S=\left\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2}=a^{2}\right\} $$ In polar coordinate system $$ \begin{aligned} &0 \leqslant r \leqslant a \\ &0 \leqslant \vartheta \leqslant 2 \pi \end{aligned} $$
So I use Stokes' theorem because the assumptions are satisfied (see below). Stokes’ theorem:
Let $D \subset \mathbb{R}^{2}$ be a regular area and let $\sigma: D \rightarrow S \subset \mathbb{R}^{3}$ be a regular oriented surface with the edge of $\partial S$ being a piecewise smooth curve. Suppose that the surface $S$ is oriented according to its parameterization, that is, in such a way that according to the right-handed screw rule, the circulation around the curve determines the return of the vector normal to the surface. Let $G \subset \mathbb{R}^{3}$ be an open set such that $G \supset S \cup \partial S$. Suppose that the function $f=(P, Q, R): G \rightarrow \mathbb{R}^{3}$ is a function of the class $C^{(1)}$ in $G$. Then:
$$\int_{\partial S} P d x+Q d y+R d z=\iint_{S}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) d y d z+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) d z d x+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y$$
Auxiliary calculation:
$$ \frac{\partial R}{\partial y}=-x\,\, \frac{\partial Q}{\partial z}=x\quad \,\,\frac{\partial P}{\partial z}=y\,\, \frac{\partial R}{\partial x}=-y\,\, \frac{\partial R}{\partial x}=z\quad \frac{\partial P}{\partial y}=z $$
Then
$$ \int_K{\underset{P}{\underbrace{\left( x^2+yz \right) }}dx+\underset{Q}{\underbrace{\left( y^2+xz \right) }}dy+\underset{R}{\underbrace{\left( z^2-xy \right) }}dz}=\iint_S{-2x\,\,dydz}+2y\,\,dxdz\,\, +\,\,0 \cdot dxdy = \ldots \star (?) $$
Question:
How to calculate this double integral ($\star$) ?
Did I make good use of Stokes' theorem?
Is there a faster way to solve this problem?
Thanks in advance.
In order to apply Stokes' theorem, $$\int_{K}\left(x^{2}+y z\right) d x+\left(y^{2}+x z\right) d y+\left(z^{2}-x y\right) d z=\iint_{S} (-2x,2y,0)\cdot d\mathbf{S}$$ a correct example of $S$ such that $\partial S=K$ is the piecewise smooth oriented surface composed by $$S_1=\{(a\cos(t),a\sin(t),\frac{st}{2\pi}, t\in [0,2\pi], s\in [0,1]\}$$ inward-oriented
$\hskip1in$
and the base disc $S_2=\{(x,y,0): x^2+y^2\leq a^2\}$ upward-oriented.
Then we find \begin{align}\iint_{S_1} (-2x,2y,0)\cdot d\mathbf{S} &=\int_{s=0}^{1}\int_{t=0}^{2\pi} (-2a\cos(t),2a\sin(t),0)\cdot (-a\cos(t) \frac{t}{2\pi},-a\sin(t) \frac{t}{2\pi},0)\,dtds\\ &=\frac{a^2}{\pi}\int_{0}^{2\pi}t(\cos^2(t)-\sin^2(t))\,dt=0 \end{align} and $$\iint_{S_2} (-2x,2y,0)\cdot d\mathbf{S} =\iint_{S_2} (-2x,2y,0)\cdot (0,0,1)\, dS=0.$$ Hence, as $S=S_1\cup S_2$, $$\iint_S (-2x,2y,0)\cdot d\mathbf{S}=\iint_{S_1} (-2x,2y,0)\cdot d\mathbf{S}+\iint_{S_2} (-2x,2y,0)\cdot d\mathbf{S}=0.$$
The same result can be obtained by direct computation of the line integral, \begin{align}\int_{K}\left(x^{2}+y z\right) d x+\left(y^{2}+x z\right) d y+\left(z^{2}-x y\right) d z&=\int_{K}-2xy\, d z\\ &=-\frac{2a^2}{2\pi}\int_{0}^{2\pi}\cos(t)\sin(t)\,dt+ \int_{1}^{0}0\,dt=0 \end{align} where we noted that the vector field $(x^2+yz,y^2+xz,z^2+xy)$ is conservative with potential function $$U=\frac{1}{3}(x^3+y^3+z^3)+xyz.$$