# Use Green's Theorem to evaluate the line integral. Orient the curve counerclockwise. oint_C F8dr, where F(x,y)=<> and C consists of the arcs y = x^2 and y = 8x for 0 <= x <= 8

Question
Use Green's Theorem to evaluate the line integral. Orient the curve counerclockwise.
$$\displaystyle\oint_{{C}}{F}{8}{d}{r}$$, where $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}$$ and C consists of the arcs $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$

2020-10-21
Step 1
We have given the line integral with values,
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}{\quad\text{and}\quad}{a}{r}{c}{s}{y}={x}^{{2}},{y}={8}{x}$$
Step 2
We know the green's theorem formula to calculate line integral,
$$\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
Now we shall plug all the values in the formula,
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{8}}}{\int_{{{x}^{{2}}}}^{{{8}{x}}}}{\left({2}{x}-{0}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{8}}}{\int_{{{x}^{{2}}}}^{{{8}{x}}}}{2}{x}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{8}}}{2}{x}{{\left[{y}\right]}_{{{x}^{{2}}}}^{{{8}{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{8}}}{2}{x}{\left[{8}{x}-{x}^{{2}}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{8}}}{{\left[{16}{x}^{{2}}-{2}{x}^{{3}}\right]}_{{0}}^{{8}}}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={{\left[{16}\frac{{x}^{{3}}}{{3}}-\frac{{{2}{x}^{{4}}}}{{4}}\right]}_{{0}}^{{8}}}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\left[{16}\frac{{8}^{{3}}}{{3}}-\frac{{8}^{{4}}}{{2}}\right]}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\left[{16}\frac{{512}}{{3}}-\frac{{4096}}{{2}}\right]}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\left[{2730.66}-{2048}\right]}$$
$$\displaystyle\int\int_{{D}}\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={682.66}$$
Step 3
So the value of line integral is 682.66

### Relevant Questions

Use Green's Theorem to evaluate F * dr. (Check the orientation of the curve before applying the theorem.)
$$\displaystyle{F}{\left({x},{y}\right)}=\sqrt{{x}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}$$
C consists of the arc of the curve $$\displaystyle{y}={\sin{{\left({x}\right)}}}$$ from (0, 0) to $$\displaystyle{\left(\pi,{0}\right)}$$ and the line segment from $$\displaystyle{\left(\pi,{0}\right)}$$ to (0, 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 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 $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
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.
Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad g · n = Dng occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of g.)
$$\displaystyle\oint_{{c}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
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).
$$\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}^{{2}}{y}{\left.{d}{y}\right.}$$
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 $$\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
$$\displaystyle\oint_{{C}}{\left({x}^{{2}}+{y}\right)}{\left.{d}{x}\right.}-{\left({3}{x}+{y}^{{3}}\right)}{\left.{d}{y}\right.}$$
Where c is the ellipse $$\displaystyle{x}^{{2}}+{4}{y}^{{2}}={4}$$