# Let F(x,y)= << 4 cos(y),2 sin(y)>>. Compute the flux oint F * nds of F across the boundary of the rectangle 0 <= x <= 5,0 <= y <= pi/2 using the vector form of Green's Theorem. oint F*nds=?

Let $F\left(x,y\right)=⟨4\mathrm{cos}\left(y\right),2\mathrm{sin}\left(y\right)⟩$. Compute the flux $\oint F\cdot nds$ of F across the boundary of the rectangle $0\le x\le 5,0\le y\le \frac{\pi }{2}$ using the vector form of Greens
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Step 1
By using Green Theorem,
$\oint F.nds={\int }_{x}{\int }_{y}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$
$={\int }_{x=0}^{5}{\int }_{y=0}^{\frac{\pi }{2}}\left(\frac{\partial \left(2\mathrm{sin}y\right)}{\partial x}-\frac{\partial \left(2\mathrm{sin}y\right)}{\partial y}\right)dxdy$
$={\int }_{x=0}^{5}{\int }_{y=0}^{\frac{\pi }{2}}\left(0-2\mathrm{cos}y\right)dxdy$
$=-10{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}ydy$
$=-10{\left(\mathrm{sin}y\right)}_{0}^{\frac{\pi }{2}}$
$=-10\left(1-0\right)$
$=-10$
Step 2
CONCLUSION:
The correct answer for the flux integral is -10.