Theorem of Green, theorem of Gauss and theorem of Stokes

We will follow the procedure:

• First we must calculate the parametrization of the surface.

$\sigma(x,y)=\Big(x,y,\dfrac{x^2+y^2}{2}\Big)$, as $z\leq2$, we have that $x^2+y^2 \leq 4$, $(x,y)$ take values inside a circle with radius $2$.

On the other hand, the curve $C$ is the circumference in $z=2$, with radius $2$, as is shown in the picture, and its parametrization will be $$\gamma(t)=(2\cdot\cos(t),2\cdot\sin(t),2), \mbox{ for } t\in[0,2\pi]$$

• We calculate $$rot(F)=\Big(\dfrac{d}{dy}F_3-\dfrac{d}{dz}F_2,\dfrac{d}{dz}F_1-\dfrac{d}{dx}F_3,\dfrac{d}{dx}F_2-\dfrac{d}{dy}F_2\Big)=$$ $$=(z^2+x,0-0,-z-3)$$

• We calculate now using our knowledge on vector analysis, $$\int_S rot(F)dS=\int_S rot(F(\sigma(x,y)))dS=$$ $$=\int_S \Big(\Big( \dfrac{x^2+y^2}{2}\Big)^2+x,0,-\dfrac{x^2+y^2}{2}-3\Big)\cdot(T_x \times T_y) \ dxdy$$ Where $Tx = (1,0, x), Ty = (0,1, y)$, and therefore, $T_x \times T_y = (-x, - y, 1)$. Then: $$\int_S rot(F)dS=-\int_S \Big(\Big( \dfrac{x^2+y^2}{2}\Big)^2\cdot x+x^2+\dfrac{x^2+y^2}{2}+3\Big) \ dxdy=$$ $$=\lbrace\mbox{Change to polar coordinates } (|J|=r)\rbrace=$$ $$=-\int_0^2\int_0^{2\pi}\Big(\dfrac{r^5}{4}\cdot\cos(t)+r^2\cdot\cos^2(t)+\dfrac{r^2}{2}+3\Big)\cdot r\cdot dtdr=$$ $$=\lbrace\mbox{ we use that } \cos^2(t)=\dfrac{1+\cos(2t)}{2}\rbrace=$$ $$-\int_0^2\int_0^{2\pi}\Big(\dfrac{r^6}{4}\cdot\cos(t)+r^3\cdot\dfrac{1+\cos(2t)}{2}+\dfrac{r^3}{2}+3r\Big)dtdr=$$ $$=\lbrace \mbox{the integral of the cosine between } 0 \mbox{ and } 2\pi \mbox{ is zero}\rbrace=$$ $$=-2\cdot\Big[\dfrac{r^4}{8}\Big]_0^2\cdot[t]_0^{2\pi}-3\Big[\dfrac{r^2}{2}\Big]_0^2\cdot[t]_0^{2\pi}=-20\pi$$

We now calculate the integral using the parametrization of the curve $C$: $\gamma(t)=(2\cdot\cos(t),2\cdot\sin(t),2), \mbox{ for } t\in[0,2\pi]$

$$\int_C F\cdot dL=\int_0^{2\pi} F(\gamma(t))\cdot \gamma'(t)dt=\int_0^{2\pi} (6\sin(t),-4\cos(t),8\sin(t))\cdot(-2\sin(t),2\cos(t),0)dt=$$ $$-4\int_0^{2\pi}(3\sin^2(t)+2\cos^2(t))dt=\left\{\begin{array}{c} 2\sin^2(t)+2\cos^2(t)=2 \\ \sin^2(t)=\dfrac{1-\cos(2t)}{2} \end{array}\right\}=$$ $$=-4\int_0^{2\pi} \Big(2+\dfrac{1-\cos(2t)}{2}\Big)dt=-8\cdot2\pi-4\cdot\dfrac{1}{2}\cdot2\pi=-20\pi$$ and therefore the Stokes theorem is satisfied.