Mylo O'Moore

2021-01-13

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

unessodopunsep

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

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