# Use Stokes' Theorem to evaluate int int_S CURL f * dS. F(x,y,z)=x^2y^3zi+sin(xyz)j+xyzk, S is the part of the cone 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.

Question
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.

2021-03-10
Solution:
The vector field is $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$
Find: $$\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}$$
Then, the double integral becomes $$\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=\int_{{C}}{F}·{d}{r}$$
The surface of a cone is $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$.
At $$\displaystyle{y}={2},{x}^{{2}}+{z}^{{2}}={4}$$.
Let the parametric equation is $$\displaystyle{r}{\left({t}\right)}={<}{2}{\cos{{t}}},{2},{2}{\sin{{t}}}{>}\forall{0}≤{t}≤{2}\pi$$ and its derivative is $$\displaystyle{r}'{\left({t}\right)}={<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}$$.
Then, the vector field becomes
$$\displaystyle{F}{\left({r}{\left({t}\right)}\right)}=={\left({2}{\cos{{t}}}\right)}^{{2}}{\left({2}\right)}^{{3}}{\left({2}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}{\left({64}{{\cos}^{{2}}{t}}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}$$
Step 2
Obtain the integal:
$$\displaystyle\int_{{C}}{F}·{d}{r}={\int_{{0}}^{{{2}\pi}}}{0}{<}{64}{{\cos}^{{2}}{t}}{\sin{{t}}},{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}},{8}{\sin{{t}}}{\cos{{t}}}{>}{<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}{\int_{{0}}^{{{2}\pi}}}{\left(−{128}{{\cos}^{{2}}{t}}{{\sin}^{{2}}{t}}+{16}{\sin{{t}}}{{\cos}^{{2}}{t}}\right)}{\left.{d}{t}\right.}−{32}\pi$$
Therefore, $$\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=−{32}\pi$$.

### Relevant Questions

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 $$\displaystyle\int{S}{c}{u}{r}{l}{F}\cdot{d}{S}{\quad\text{if}\quad}{F}={y}{i}+{z}{j}+{2}{x}{k}$$.
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.
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 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.
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).
Consider the vector field $$\displaystyle{F}={\left\langle{5}{z},{x},{5}{y}\right\rangle}$$ and the surface which is the part of the elliptic paraboloid $$\displaystyle{z}={x}^{{2}}+{5}{y}^{{2}}$$ that lies below the plane z = 5. Calculate curl(F) and then apply Stokes' Theorem to compute the exact magnitude of the flux of curl(F) through the surface using line integral. You do not need to cinfirm your answer by evaluating the double integral of curl(F) over the surface(the right-hand side of Stokes' Theorem).
Use the divergence theorem to evaluate $$\displaystyle\int\int_{{S}}{F}\cdot{N}{d}{S}$$, where $$\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}$$ and S is the boundary of the cube defined by $$\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
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.
If E(t,x,y,z) and B(t,x,y,z)represent the electric and magnetic fields at point (x,y,z) at time t, a basic principle of electromagnetic theory says that $$\displaystyle\nabla\times{E}=\frac{{-\partial{B}}}{{\partial{t}}}$$. In this expression $$\displaystyle\nabla\times{E}$$ is computed with t held fixed and $$\displaystyle\frac{{\partial{B}}}{{\partial{t}}}$$ is calculated with (x,y,z) fixed.
Use Stokes' Theorem to derive Faraday's law, $$\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma$$,
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.