# 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 S curl F* dS if F = yi + zj + 2xk.

Question
z = x Let be the curve of intersection of the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ 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 $$\displaystyle\int{S}{c}{u}{r}{l}{F}\cdot{d}{S}{\quad\text{if}\quad}{F}={y}{i}+{z}{j}+{2}{x}{k}$$.

2021-03-09
Step 1
Stock's Theorem:-
$$\displaystyle\int_{{C}}{F}.{d}{r}\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}$$
Given that a curve say D (as name not mention in problem) of intersection of the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the plane z=x and F=y i+z j+ 2x k
Also, S be inside of this curve, oriented with upward-pointing normal, parameterize of this curve is
$$\displaystyle{r}={<}{x},{y},{z}\ge{<}{x},{y},{x}{>}$$ as z = z
$$\displaystyle{r}={\cos{{0}}}{i}+{\sin{{0}}}{j}+{\cos{{0}}}{k}$$
since, we know $$\displaystyle{x}={1}.{\cos{{0}}}={\cos{{0}}},{y}={1}.{\sin{{0}}}={\sin{{0}}},{z}={x}={\cos{{0}}}$$ as radius =1
Here, $$\displaystyle{0}\le{0}\le{2}\pi{\left({a}{s}{x}^{{2}}+{y}^{{2}}={1}^{{2}}\right)}$$
then,
$$\displaystyle{r}={<}{\cos{{0}}},{\sin{{0}}},{\cos{{0}}}{>}$$
$$\displaystyle\frac{{{d}{r}}}{{{d}{0}}}={r}'={<}-{\sin{{0}}},{\cos{{0}}},-{\sin{{0}}}{>}$$
F(x,y,z)=yi+zj+2xk
$$\displaystyle{F}{\left({r}{\left({0}\right)}\right)}={\sin{{0}}}{i}+{\cos{{0}}}{j}+{2}{\cos{{0}}}{k}$$
Step 2
Now,
$$\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{F}{\left({r}{\left({0}\right)}\right)}\frac{{{d}{r}}}{{{d}{0}}}{d}{0}$$
$$\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{<}{\sin{{0}}},{\cos{{0}}},{2}{\cos{{0}}}{>}{<}-{\sin{{0}}},{\cos{{0}}},-{\sin{{0}}}{>}{d}{0}$$
$$\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{\left(-{{\sin}^{{2}}{0}}+{{\cos}^{{2}}{0}}-{2}{\sin{{0}}}{\cos{{0}}}\right)}{d}{0}$$
Note: $$\displaystyle-{{\cos}^{{2}}{0}}-{{\sin}^{{2}}{0}}={\cos{{20}}},{\sin{{20}}}={2}{\sin{{0}}}{\cos{{0}}}$$
$$\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{\left({\cos{{20}}}-{\sin{{20}}}\right)}{d}{0}={{\left[\frac{{{\sin{{20}}}}}{{2}}+\frac{{{\cos{{20}}}}}{{2}}\right]}_{{0}}^{{{2}\pi}}}$$
Note: $$\displaystyle-{\sin{{\left({2}\pi\right)}}}={\sin{{\left({0}\right)}}}={0},{\cos{{\left({2}\pi\right)}}}-{\cos{{\left({0}\right)}}}={1}-{1}={0}$$
$$\displaystyle\int_{{D}}{F}{d}{r}={0}$$
By, using Stock's Theorem, we have
$$\displaystyle\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}=\int_{{D}}{F}.{d}{r}={0}$$
$$\displaystyle\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}={0}$$

### Relevant Questions

Use Stokes' Theorem to evaluate int_C F*dr where C is oriented counterclockwise as viewed above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.
Use Stokes' Theorem to evaluate $$\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$,
S is the part of the cone $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
C is the triangle with vertices (3,0,0),(0,3,0), and (0,0,3).
Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
S is the surface of the solid bounded by the cylinder $$\displaystyle{y}^{{2}}+{z}^{{2}}={9}$$ and the planes x = −3 and x = 1.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Use Stokes' theorem to evaluate the line integral $$\displaystyle\oint_{{C}}{F}\cdot{d}{r}$$ where A = -yi + xj and C is the boundary of the ellipse $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem $$\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
Double integral $$\displaystyle{F}\cdot{a}{S}$$ for each of the following regions W:
a. $$\displaystyle{x}^{{2}}+{y}^{{2}}$$ (less than or equal to) z (less thanor equal to) 1
b. $$\displaystyle{x}^{{2}}+{y}^{{2}}$$ (less than or equal to) 1 and x (greater than or equal to) 0