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 * 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)

Question
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)

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.
\(\displaystyle\oint_{{C}}{F}\cdot{d}{r}={\int_{{0}}^{\pi}}{\int_{{0}}^{{{\sin{{x}}}}}}{\left(-{8}{x}+{12}{y}^{{2}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{{\left(-{8}{x}{y}+\frac{{{12}{y}^{{3}}}}{{3}}\right)}_{{0}}^{{{\sin{{x}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\left(-{8}{x}{\left({\sin{{x}}}\right)}+{4}{{\sin}^{{3}}{x}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\int_{{0}}^{\pi}}{8}{x}{\sin{{x}}}{\left.{d}{x}\right.}+{\int_{{0}}^{\pi}}{4}{{\sin}^{{3}}{x}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}{{\left({x}{\left(-{\cos{{x}}}\right)}-\int{\cos{{x}}}{\left.{d}{x}\right.}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{{\sin}^{{2}}{x}}{\sin{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}{{\left(-{x}{\cos{{x}}}-{\sin{{x}}}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{\left({1}-{{\cos}^{{2}}{x}}\right)}{\sin{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{8}\pi+{4}{{\left(-{\cos{{x}}}-\frac{{{{\cos}^{{3}}{x}}}}{{3}}\right)}_{{0}}^{\pi}}\)
\(\displaystyle=-{8}\pi+{4}{\left({2}-\frac{{2}}{{3}}\right)}\)
\(\displaystyle=-{8}\pi+\frac{{16}}{{3}}\)
0

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