# Let F(x,y)=<<xy^2+8x, x^2y-8y>>. Compute the flux oint F * nds of F across a simple closed curve that is the boundary of the half-disk given by x^2+y^2<=7, y>= 0 using the vector form Green's Theorem. oint F * nds = ?

Question
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}{y}^{{2}}+{8}{x},{x}^{{2}}{y}-{8}{y}\right\rangle}$$. Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across a simple closed curve that is the boundary of the half-disk given by $$\displaystyle{x}^{{2}}+{y}^{{2}}\le{7},{y}\ge{0}$$ using the vector form Green's Theorem.
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$ ?

2020-12-18
Step 1
Consider the equation
Step 2
$$\displaystyle{F}{\left({x},{y}\right)}={\left({x}{y}^{{2}}+{8}{x},{x}^{{2}}{y}-{8}{y}\right)}$$
$$\displaystyle\oint{F}.{n}{d}{s}==\int\int\nabla\times{D}{d}{A}$$
$$\displaystyle\oint{F}.{n}{d}{s}=\int\int_{{S}}{\left({8}{x}{\left.{d}{y}\right.}+{y}\right)}={3}\pi$$
$$\displaystyle\int{\left({8}{x},{y}\right)}.{\left({x},{y}\right)}{d}{s}=\int{8}{x}{\left.{d}{y}\right.}+{y}{\left(-{\left.{d}{x}\right.}\right)}$$
$$\displaystyle{x}={\cos{{0}}}{y}={\sin{{0}}}{d}{s}={r}{d}{0}\Rightarrow{d}{s}={d}{0}$$ because the radius is 1
$$\displaystyle{\left.{d}{x}\right.}=-{y}{d}{0}{\left.{d}{y}\right.}={\cos{{0}}}{d}{0}$$
$$\displaystyle\oint{\left(-{y},{8}{x}\right)}.{\left.{d}{x}\right.}$$
$$\displaystyle\int\int_{{S}}\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}={3}$$
(the area) $$\displaystyle={3}{\left(\pi{\left({1}\right)}^{{2}}\right)}={3}\pi$$

### Relevant Questions

Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{4}{\cos{{\left({y}\right)}}},{2}{\sin{{\left({y}\right)}}}\right\rangle}$$. Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$?
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{5}{\cos{{\left({y}\right)}}},{8}{\sin{{\left({y}\right)}}}\right\rangle}$$.Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
(Use symbolic notation and fractions where needed.)
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$ ?
True or false?
Given an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve with positive orientation, and a vector field whose components have continuous partial derivatives on an open region in three-dimensional Euclidean space. Then the line integral of this vector field along the boundary curve equals the flux of the curl of the vector field across the surface.
Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Divergence Theorem for more general regions Use the Divergence Theorem to compute the net outward flux of the following vector fields across the boundary of the given regions D.
$$\displaystyle{F}={\left\langle{z}-{x},{x}-{y},{2}{y}-{z}\right\rangle}$$, D is the region between the spheres of radius 2 and 4 centered at the origin.
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.
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 $$\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
where n(t) is the outward unit normal vector to C.
$$\displaystyle{F}={\left\langle{x},{y}\right\rangle},{R}={\left\lbrace{\left({x},{y}\right)}:{x}^{{2}}+{y}^{{2}}\le{4}\right\rbrace}$$
Let $$\displaystyle{f}={\left[{x}^{{2}}{y}^{{2}},-\frac{{x}}{{y}^{{2}}}\right]}$$ and $$\displaystyle{R}:{1}\le{x}^{{2}}+{y}^{{2}},+{4},{x}\ge{0},{y}\ge{x}$$. Evaluate $$\displaystyle\int_{{C}}{F}{\left({r}\right)}\cdot{d}{r}$$ counterclockwise around the boundary C of the region R by Green's theorem.