CMIIh

2021-03-08

z = x Let be the curve of intersection of the cylinder ${x}^{2}+{y}^{2}=1$ and the plane , oriented positively when viewed from above . Let S be the inside of this curve , oriented with upward -pointing normal . Use Stokes ' Theorem to evaluate $\int ScurlF\cdot dS\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}F=yi+zj+2xk$.

lamanocornudaW

Step 1
Stock's Theorem:-
${\int }_{C}F.dr\int \int curl\left(F\right)ds$
Given that a curve say D (as name not mention in problem) of intersection of the cylinder ${x}^{2}+{y}^{2}=1$ and the plane z=x and
Also, S be inside of this curve, oriented with upward-pointing normal, parameterize of this curve is
$r=$ as z = z
$r=\mathrm{cos}0i+\mathrm{sin}0j+\mathrm{cos}0k$
since, we know $x=1.\mathrm{cos}0=\mathrm{cos}0,y=1.\mathrm{sin}0=\mathrm{sin}0,z=x=\mathrm{cos}0$ as radius =1
Here, $0\le 0\le 2\pi \left(as{x}^{2}+{y}^{2}={1}^{2}\right)$
then,
$r=<\mathrm{cos}0,\mathrm{sin}0,\mathrm{cos}0>$
$\frac{dr}{d0}={r}^{\prime }=<-\mathrm{sin}0,\mathrm{cos}0,-\mathrm{sin}0>$
$F\left(x,y,z\right)=yi+zj+2xk$
$F\left(r\left(0\right)\right)=\mathrm{sin}0i+\mathrm{cos}0j+2\mathrm{cos}0k$
Step 2
Now,
${\int }_{D}Fdr={\int }_{0}^{2\pi }F\left(r\left(0\right)\right)\frac{dr}{d0}d0$
${\int }_{D}Fdr={\int }_{0}^{2\pi }<\mathrm{sin}0,\mathrm{cos}0,2\mathrm{cos}0><-\mathrm{sin}0,\mathrm{cos}0,-\mathrm{sin}0>d0$
${\int }_{D}Fdr={\int }_{0}^{2\pi }\left(-{\mathrm{sin}}^{2}0+{\mathrm{cos}}^{2}0-2\mathrm{sin}0\mathrm{cos}0\right)d0$
Note: $-{\mathrm{cos}}^{2}0-{\mathrm{sin}}^{2}0=\mathrm{cos}20,\mathrm{sin}20=2\mathrm{sin}0\mathrm{cos}0$
${\int }_{D}Fdr={\int }_{0}^{2\pi }\left(\mathrm{cos}20-\mathrm{sin}20\right)d0={\left[\frac{\mathrm{sin}20}{2}+\frac{\mathrm{cos}20}{2}\right]}_{0}^{2\pi }$
Note:

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