Question

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

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