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 \(\displaystyle\frac{\pi}{{2}}\).

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 \(\displaystyle\frac{\pi}{{2}}\).