Problems from Theorem of Green, theorem of Gauss and theorem of Stokes

Development:

We will follow the procedure:

• First we must calculate the parametrization of the surface.

$\sigma (x,y)=(x,y,{\displaystyle \frac{x^2+y^2}{2}})$, as $z\le 2$, we have that $x^2+y^2\le 4$, $(x,y)$ 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 $$\gamma (t)=(2\cdot \cos (t),2\cdot \sin (t),2),\text{ for }t\in [0,2\pi ]$$

• We calculate $$rot(F)=({\displaystyle \frac{d}{dy}}{F}_3-{\displaystyle \frac{d}{dz}}{F}_2,{\displaystyle \frac{d}{dz}}{F}_1-{\displaystyle \frac{d}{dx}}{F}_3,{\displaystyle \frac{d}{dx}}{F}_2-{\displaystyle \frac{d}{dy}}{F}_2)=$$ $$=(z^2+x,0-0,-z-3)$$

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

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

$${\int }_{C}F\cdot dL={\int }_0^{2\pi }F(\gamma (t))\cdot {\gamma }^{\prime }(t)dt={\int }_0^{2\pi }(6\sin (t),-4\cos (t),8\sin (t))\cdot (-2\sin (t),2\cos (t),0)dt=$$ $$-4{\int }_0^{2\pi }(3{\sin }^2(t)+2{\cos }^2(t))dt=\left\{\begin{array}{c}2{\sin }^2(t)+2{\cos }^2(t)=2\\ {\sin }^2(t)={\displaystyle \frac{1-\cos (2t)}{2}}\end{array}\right\}=$$ $$=-4{\int }_0^{2\pi }(2+{\displaystyle \frac{1-\cos (2t)}{2}})dt=-8\cdot 2\pi -4\cdot {\displaystyle \frac{1}{2}}\cdot 2\pi =-20\pi$$ and therefore the Stokes theorem is satisfied.

Solution:

The Stokes theorem is satisfied.

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