Question

# 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 $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ from(0,1)to(1,0).

2021-01-06
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 $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$
from(0,1)to(1,0)
Step 2
The work done can be calculated as:
work done = $$\displaystyle\int_{{C}}{F}.{d}{r}$$
Step 3
Solving further to get:
$$\displaystyle\int_{{C}}{F}.{d}{r}$$
$$\displaystyle{r}{\left({t}\right)}={\left({\sin{{t}}}\right)}{i}+{\left({\cos{{t}}}\right)}{j}{f}{\quad\text{or}\quad}{0}\le{t}\le\frac{\pi}{{2}}$$
$$\displaystyle{r}'{\left({t}\right)}={\left({\cos{{t}}}\right)}{i}+{\left(-{\sin{{t}}}\right)}{j}$$
$$\displaystyle\int_{{C}}{F}.{d}{r}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({\left({4}{\cos{{t}}}\right)}{i}+{\left({2}{\sin{{t}}}\right)}{j}\right)}.{\left({\left({\cos{{t}}}\right)}{i}+{\left(-{\sin{{t}}}\right)}{j}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({4}{{\cos}^{{2}}{t}}-{2}{{\sin}^{{2}}{t}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({6}{{\cos}^{{2}}{t}}-{2}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=\frac{\pi}{{2}}$$
Therefore, the total work done by the force field is $$\displaystyle\frac{\pi}{{2}}$$.