# Apply Green’s theorem to find the outward flux for the field F(x,y)=tan^(−1)(y/x)i+ln(x^2+y^2)j

Question
Apply Green’s theorem to find the outward flux for the field
$$\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{−{1}}}{\left(\frac{{y}}{{x}}\right)}}{i}+{\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}{j}$$

2021-03-06
Step 1
We are given that $$\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{i}+{\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}{j}$$ and the curve C defined as the boundary of the given region.
We will be using Green's theorem to find the outward Flux.
Let $$\displaystyle{P}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{\quad\text{and}\quad}{Q}{\left({x},{y}\right)}={\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}$$.
Using the first version of Green's theorem, we get
Flux = $$\displaystyle\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$.
Step 2
Now,
$$\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}}}{2}{x}=\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}$$
And $$\displaystyle\frac{{\partial{P}}}{{\partial{y}}}=\frac{{1}}{{x}}\frac{{1}}{{{1}+{\left(\frac{{y}}{{x}}\right)}^{{2}}}}=\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}$$.
So, Flux = $$\displaystyle\int\int_{{R}}{\left(\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}-\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
In polar form, $$\displaystyle{x}={r}{\cos{{0}}}{\quad\text{and}\quad}{y}={r}{\sin{{0}}}$$.
Jacobean = $$\displaystyle{\left[\begin{array}{cc} {\cos}&{\sin{{0}}}\\-{r}{\sin{{0}}}&{r}{\cos{{0}}}\end{array}\right]}={r}$$.
Thus, Flux = $$\displaystyle\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}\frac{{{r}{\cos{{0}}}}}{{r}^{{2}}}{r}{d}{r}{d}{0}$$
$$\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}{\cos{{0}}}{d}{r}{d}{0}$$
$$\displaystyle={\int_{{0}}^{\pi}}{\cos{{0}}}{d}{0}$$
$$\displaystyle={{\left[{\sin{{0}}}\right]}_{{0}}^{\pi}}$$
=0

### Relevant Questions

use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. $$\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}$$
C: The triangle bounded by y = 0, x = 3, and y = x
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. F = (x - y)i + ( y - x)j
C: The square bounded by x = 0, x = 1, y = 0, and y = 1
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.
Green’s Theorem, flux form Consider the following regions R and vector fields F.
a. Compute the two-dimensional divergence of the vector field.
b. Evaluate both integrals in Green’s Theorem and check for consistency.
$$\displaystyle{F}={\left\langle{x},{y}\right\rangle},{R}={\left\lbrace{\left({x},{y}\right)}:{x}^{{2}}+{y}^{{2}}\le{4}\right\rbrace}$$
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.
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).
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.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}$$
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.
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.