 # Using Green's theorem evaluate oint_C(6y+x)dx+(y+2x)dy where C is the boundary Cheyanne Leigh 2020-11-12 Answered

Using Green's theorem evaluate ${\oint }_{C}\left(6y+x\right)dx+\left(y+2x\right)dy$ where C is the boundary of the

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it avortarF
Here we have to evalute the integral $\int \left(6y+x\right)dx+\left(y+2x\right)dy$ over the regin bounced by $y={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x={y}^{2}$ say the regin D.By Greens theorem
$\int \left(6y+x\right)dx+\left(y+2x\right)dy=\int {\int }_{D}\left(\frac{\partial }{\partial x}\left(y+2x\right)-\frac{\partial }{\partial y}\right)\left(6y+x\right)dydx$
${\int }_{0}^{1}{\int }_{y={x}^{2}}^{y=\sqrt{x}}\left(\frac{\partial }{\partial x}\left(y+2x\right)-\frac{\partial }{\partial y}\left(6y+x\right)\right)dydx$
$={\int }_{0}^{1}{\int }_{y={x}^{2}}^{y=\sqrt{x}}\left(2-6\right)dydx$
$-4{\int }_{0}^{1}{\int }_{y={x}^{2}}^{y=\sqrt{x}}dydx$
$-4{\int }_{0}^{1}\left(\sqrt{x}-{x}^{2}\right)dx$
$=-4\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{x}^{3}}{3}{|}_{0}^{1}$ $=-4\left(\frac{2}{3}-\frac{1}{3}\right)$
$=-\frac{4}{3}$