# Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3). Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle x^2+y^2=1 from(0,1)to(1,0).

Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3).
Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle ${x}^{2}+{y}^{2}=1$ from(0,1)to(1,0).
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Step 1
According to the given information, it is required to calculate the work done by the force field
F(x,y)=4yi + 2xj
moving along the particle along a circle ${x}^{2}+{y}^{2}=1$
from(0,1)to(1,0)
Step 2
The work done can be calculated as:
work done = ${\int }_{C}F.dr$
Step 3
Solving further to get:
${\int }_{C}F.dr$
$r\left(t\right)=\left(\mathrm{sin}t\right)i+\left(\mathrm{cos}t\right)jf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}0\le t\le \frac{\pi }{2}$
${r}^{\prime }\left(t\right)=\left(\mathrm{cos}t\right)i+\left(-\mathrm{sin}t\right)j$
${\int }_{C}F.dr={\int }_{0}^{\frac{\pi }{2}}\left(\left(4\mathrm{cos}t\right)i+\left(2\mathrm{sin}t\right)j\right).\left(\left(\mathrm{cos}t\right)i+\left(-\mathrm{sin}t\right)j\right)dt$
$={\int }_{0}^{\frac{\pi }{2}}\left(4{\mathrm{cos}}^{2}t-2{\mathrm{sin}}^{2}t\right)dt$
$={\int }_{0}^{\frac{\pi }{2}}\left(6{\mathrm{cos}}^{2}t-2\right)dt$
$=\frac{\pi }{2}$
Therefore, the total work done by the force field is $\frac{\pi }{2}$.