$F(x,y,z)=yj-zk,S$ consists of the paraboloid $y=x^2+z^2,0y1,$ and the disk $x^2+z^{2}1,y=1$

$F(x,y,z)=yj-zk.$ Let $S_1$ be the paraboloid $y=x^2+z^2,0y1$ and $S_2$ the disk $x^2+z^{2}1,y=1.$ On $S_1$ we have $r(x,z)=+(x^2+z^2)j+zk.$ Since $S$ is a closed surface, we use the outward orientation.

On $S_1$:$F(r(x,z))=(x^2+z^2)j-zk$ and $r_x{r}_z=2-j+2zk($since the $j-$component must be negative on $S_1).$

Then by Formula $9,$

${}_{S_1}F*dS={}_{x^2+z^2}1[0-(x^2+z^2)-2z^2]dA=-{\displaystyle \underset{0}{\overset{2}{}}}{\displaystyle \underset{0}{\overset{1}{}}}(r^2+2r^{2}si{n}^2)rdrd$

$=-{\displaystyle \underset{0}{\overset{2}{}}}{\displaystyle \underset{0}{\overset{1}{}}}{r}^3(1+2si{n}^2)drd=-{\displaystyle \underset{0}{\overset{2}{}}}(1+1-cos2)d{\displaystyle \underset{0}{\overset{1}{}}}{r}^{3}dr$

$=-[2-\frac{1}{2}sin2{]}_0^2[\frac{1}{4}{r}^4{]}_0^1=-4*\frac{1}{4}=-$On $S_2$:$F(r(x,z))=j-zk$ and $r_z{r}_x=j.$ Then ${}_{S_2}F*dS={}_{x^2+z^2}1(1)dA=.$ Hence ${}_{S}F*dS=-+=0.$a$----------------------------------------$Why is it $\text{'}$r$\_$x crossr$\_$z$\text{'}$ at the top and $\text{'}$r$\_$z crossr$\_$x$\text{'}$ at the bottom? Also, what doe

s it mean when it says $\text{'}$j$-$component must be negative'? Please let me know kindly.

$21-32$ Evaluate the surface integral ${}_{S}F*dS$ for the given vector field $F$ and the oriented surface $S.$ In other words, f ind the flux of $F$ across $S.$ For closed surfaces, use the positive $($outward$)$ orientation.