# Use Green's Theorem to find int_C vec(F) * d vec(r) where vec(F) = <> and C is the circle x^2+y^2=3.

Question
Use Green's Theorem to find $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ where $$\displaystyle\vec{{{F}}}={\left\langle{y}^{{3}},-{x}^{{3}}\right\rangle}$$ and C is the circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={3}$$.

2020-11-09
Step 1
The given vector is
$$\displaystyle{F}={y}^{{3}}{i}-{x}^{{3}}{j}$$
Step 2
$$\displaystyle{I}=\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$
but d $$\displaystyle\vec{{{r}}}={\left.{d}{x}\right.}\hat{{{i}}}+{\left.{d}{y}\right.}\hat{{{j}}}$$
$$\displaystyle{I}=\int_{{C}}{\left({y}^{{3}}\hat{{{i}}}-{x}^{{3}}\hat{{{j}}}\right)}.{\left({\left.{d}{x}\right.}\hat{{{i}}}+{\left.{d}{y}\right.}\hat{{{j}}}\right)}$$
$$\displaystyle=\int_{{C}}{\left({y}^{{3}}{\left.{d}{x}\right.}-{x}^{{3}}{\left.{d}{y}\right.}\right)}$$
The given circle
$$\displaystyle{x}^{{2}}+{y}^{{2}}={3}$$
$$\displaystyle{x}^{{2}}+{y}^{{2}}={\left(\sqrt{{{3}}}\right)}^{{2}}$$
$$\displaystyle{r}{a}{d}{i}{u}{s}=\sqrt{{{3}}}$$
Apply Green Theorem
$$\displaystyle\int_{{C}}{\left({M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}\right)}=\int\int{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{{D}{e}{l}{M}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle\Rightarrow{I}=\int\int{\left(-{3}{x}^{{2}}-{3}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=\int\int-{3}{\left({x}^{{2}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=-{3}\int\int{\left({x}^{{2}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=-{9}\int\int{\left.{d}{x}\right.}{\left.{d}{y}\right.}\because{x}^{{2}}+{y}^{{2}}={3}$$
=-9 x Area of circle
$$\displaystyle=-{9}{x}\pi{r}^{{2}}$$
$$\displaystyle=-{9}{x}\pi{x}{\left(\sqrt{{{3}}}\right)}^{{2}}$$
$$\displaystyle{I}=-{27}\pi$$

### Relevant Questions

Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ where $$\displaystyle\vec{{{F}}}{\left({x},{y}\right)}={x}{y}^{{2}}{i}+{\left({1}-{x}{y}^{{3}}\right)}{j}$$ and C is the parallelogram with vertices (-1,2), (-1,-1),(1,1)and(1,4).
The orientation of C is counterclockwise.
Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3).
Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ from(0,1)to(1,0).
Use Green's Theorem to evaluate the line integral
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$
where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise.
a)6
b)10
c)14
d)4
e)8
f)12
Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}$$ where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}$$.
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).
Evaluate $$\displaystyle\int_{{C}}{x}^{{2}}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}{y}^{{3}}{\left.{d}{y}\right.}$$ where C is the triangle with vertices(0,0),(1,3), and (0,3).
(a)Use the Green's Theorem.
(b)Do not use the Green's Theorem.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.
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.
$$\displaystyle\oint_{{c}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
$$\displaystyle\oint_{{C}}{F}{8}{d}{r}$$, where $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}$$ and C consists of the arcs $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$