Question

Use Green's Theorem to evaluate F * dr. (Check the orientation of the curve before applying the theorem.)F(x, y) = sqrtx+ 4y^3, 4x^2 + sqrt(y)C consists of the arc of the curve y = sin(x) from (0, 0) to (pi, 0) and the line segment from (pi, 0) to (0, 0)

Use Green's Theorem to evaluate \(F \cdot dr\). (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)

Expert Answers (1)

2021-01-14

Step 1
The given function is,
\(\displaystyle{F}{\left({x},{y}\right)}={\left(\sqrt{{{x}}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}\right)}\)
Step 2
C is a closed curve and using Green’s theorem for clockwise orientation the integral is evaluated using the below formula.
\(\displaystyle\oint_{{C}}{F}\cdot{d}{r}=-\int\int_{{D}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}\)
Where, \(\displaystyle{F}={\left({P},{Q}\right)}={\left(\sqrt{{{x}}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}\right)}\)
Step 3
Calculate the partial derivative of P and Q as follows.
\(\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}=\frac{{\partial}}{{\partial{x}}}{\left({4}{x}^{{2}}+\sqrt{{{y}}}\right)}\)
\(=8x\)
\(\displaystyle\frac{{\partial{P}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left(\sqrt{{{x}}}+{4}{y}^{{3}}\right)}\)
\(\displaystyle={12}{y}^{{2}}\)
Step 4
Evaluate the integral Fdr by substituting the limits as follows.
\(\displaystyle\oint_{{C}}{F}\cdot{d}{r}={\int_{{0}}^{\pi}}{\int_{{0}}^{{{\sin{{x}}}}}}{\left(-{8}{x}+{12}{y}^{{2}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{{\left(-{8}{x}{y}+\frac{{{12}{y}^{{3}}}}{{3}}\right)}_{{0}}^{{{\sin{{x}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\left(-{8}{x}{\left({\sin{{x}}}\right)}+{4}{{\sin}^{{3}}{x}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\int_{{0}}^{\pi}}{8}{x}{\sin{{x}}}{\left.{d}{x}\right.}+{\int_{{0}}^{\pi}}{4}{{\sin}^{{3}}{x}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}{{\left({x}{\left(-{\cos{{x}}}\right)}-\int{\cos{{x}}}{\left.{d}{x}\right.}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{{\sin}^{{2}}{x}}{\sin{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}{{\left(-{x}{\cos{{x}}}-{\sin{{x}}}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{\left({1}-{{\cos}^{{2}}{x}}\right)}{\sin{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}\pi+{4}{{\left(-{\cos{{x}}}-\frac{{{{\cos}^{{3}}{x}}}}{{3}}\right)}_{{0}}^{\pi}}\)
\(\displaystyle=-{8}\pi+{4}{\left({2}-\frac{{2}}{{3}}\right)}\)
\(\displaystyle=-{8}\pi+\frac{{16}}{{3}}\)

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