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Evaluate by Green's theorem \[\int_C^{\ }\ \left[\left(3x^2-8y^2\right)\ dx+\left(4y-6xy\right)\ dy\right]\] where C is the boundary of the region bounded by \[y=\sqrt{x},\ \ y=x^2\]
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$ \displaystyle F.dr=\ \left(3x^2-8y^2\right)dx+\left(4y-6xy\right)dy….\left(1\right) \\[2ex] \displaystyle Let\ P=3x^2-8y^2\ and\ Q=4y-6xy\ \\[2ex] \displaystyle \therefore{}\frac{\partial{}P}{\partial{}y}=-16y\ \ and\ \frac{\partial{}Q}{\partial{}x}=\ -6y \\[4ex] \displaystyle By\ Green’s\ Theorem,\ \\[3ex] \displaystyle \int_cPdx+Qdy=\iint_R\left(\frac{\partial{}Q}{\partial{}x}-\frac{\partial{}P}{\partial{}y}\right)dx\ dy \\[2ex] $

$\displaystyle \int_c\left(3x^2-8y^2\right)dx+\left(4y-6xy\right)dy=\iint_R\left(-6y+16y\right)dx\ \displaystyle dy…(2)\\[3ex]From\ 1\ and\ 2,\ \ \int_cF.dr=\ \iint_R10y\ dx\ dy \\[3ex] Since \ we\ need \ to \ integrate \ w.r.t \ x \ first,\\[2ex] \displaystyle y=\sqrt{x}\ \ \ \ gives\ x=y^2\ \\[2ex] \displaystyle y=x^2\ \ \ gives\ x=+\sqrt{y} \\[2ex] Also, \sqrt{x}=x^2,\ \ \ \ \ gives\ \ \ x=0,1 \\[2ex] \displaystyle when\ x=0,\ y=0,\ \ \ \ \ when\ x=1,\ y=1 \\[2ex] So,\ the \ limits\ for\ y \ are \ from \ 0\ to\ 1.\\[2ex] $

$ \begin{align*} \displaystyle \int_cF.dr&=\ 10\ \int_{y=0}^{y=1\ }y\ \left[\int_{x=y^2}^{x=\sqrt{y}}dx\ \right]\ dy\\[2ex] &=10\ \int_{y=0}^{y=1}y{\left[x\right]}_{y^2}^{\sqrt{y}}dy \\[2ex] \displaystyle &=\ 10\ \int_{y=0}^{y=1}y[\sqrt{y}-y^2]dy\ \\[2ex] \displaystyle &=10\ \int_0^1{(y}^{\frac{3}{2}}-y^3)dy \\[2ex] \displaystyle &=10\ {\left[\frac{y^{\frac{5}{2}}}{\frac{5}{2}}-\frac{y^4}{4}\right]}_0^1\\[2ex] &=10\ \left[\frac{2}{5}-\frac{1}{4}\right]\\[2ex] &=4-\frac{5}{2} \\[2ex] \displaystyle &=\frac{3}{2} \\[2ex] \end{align*}$

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