# Using Green's theorem evaluate displaystyleoint_{{C}}{left({6}{y}+{x}right)}{left.{d}{x}right.}+{left({y}+{2}{x}right)}{left.{d}{y}right.} where C is the boundary of the

Question
Multivariable functions
Using Green's theorem evaluate $$\displaystyle\oint_{{C}}{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}$$ where C is the boundary of the

2020-11-13
Here we have to evalute the integral $$\displaystyle\int{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}$$ over the regin bounced by $$\displaystyle{y}={x}^{2}{\quad\text{and}\quad}{x}={y}^{2}$$ say the regin D.By Greens theorem
$$\displaystyle\int{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}=\int\int_{{D}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}\right)}{\left({6}{y}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({6}{y}+{x}\right)}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left({2}-{6}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle-{4}{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle-{4}{\int_{{0}}^{{1}}}{\left(\sqrt{{x}}-{x}^{2}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=-{4}\frac{{{x}^{{\frac{3}{{2}}}}}}{{\frac{3}{{2}}}}-\frac{{{x}^{3}}}{{3}}{{|}_{{0}}^{{1}}}$$ $$\displaystyle=-{4}{\left(\frac{2}{{3}}-\frac{1}{{3}}\right)}$$
$$\displaystyle=-\frac{4}{{3}}$$

### Relevant Questions

a) Find the function's domain .
b) Find the function's range.
c) Find the boundary of the function's domain.
d) Determine if the domain is an open region, a closed region , or neither.
e) Decide if the domain is bounded or unbounded.
for: $$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{{{\left({16}-{x}^{{2}}-{y}^{{2}}\right)}^{{\frac{{1}}{{2}}}}}}$$
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 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
Evaluate the line integral $$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}$$, where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Green's theorem.
Write the integral(s), but do not evaluate.
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.
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).
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.
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$ ?
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.
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 $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.