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 \(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}\)

Answers (1)

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
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