91Ó°ÊÓ

The given force vector is $F=x^{2}yi-x{y}^{2}j$ and path from (1,1) to (4,2) .

Work done by a force is defined as the product of displacement of an object and component of force applied which is in the direction of displacement of an object.

The formula to find the work done by force is given by equation mentioned below.

$W=∫F.dr$

where F: force applied anddisplacement of an object.

Also, the equation of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given below.

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

Use the formula $W=∫F.dr$

Put $F=x^{2}yi-x{y}^{2}j$

and $dr=dxi+dyj+dzk$

$\begin{array}{rcl}W& =& ∫\left(x^{2}yi-x{y}^{2}j\right).\left(dxi+dyj+dyk\right)\\ W& =& ∫x^{2}ydx-x{y}^{2}dy\end{array}$

The equation of line joining points (1,1) and (4,2) is mentioned below.

$\begin{array}{rcl}\frac{y-1}{x-1}& =& \frac{2-1}{4-1}\\ \frac{y-1}{x-1}& =& \frac{1}{3}\\ x& =& 3y-2\\ dx& =& 3dy\end{array}$

Work done given below:

$\begin{array}{rcl}W& =& \underset{1}{\overset{2}{∫}}{\left(3y-2\right)}^{2}y\left(3dy\right)-\left(3y-2\right)y^{2}dy\\ W& =& \underset{1}{\overset{2}{∫}}(27y^3-36y^2+12y-3y^3+2y^2)dy\\ W& =& \underset{1}{\overset{2}{∫}}\left(24y^3-34y^2+12y\right)dy\\ W& =& \frac{86}{3}\end{array}$

(i) From (1,1) to (1,0) , we have that x = 1

Then dx = 0

$\begin{array}{rcl}{W}_1& =& \underset{1}{\overset{0}{∫}}{\left(1\right)}^{2}y\left(0\right)-\left(1\right)y^{2}dy\\ W_1& =& \underset{1}{\overset{0}{∫}}-y^{2}dy\\ W_1& =& \frac{1}{3}\end{array}$

(ii) From (1,0) to (4,0) , we have that y = 0

Then dy = 0

$\begin{array}{rcl}{W}_2& =& \underset{1}{\overset{4}{∫}}{\left(x\right)}^2\left(0\right)dx-x{\left(0\right)}^2\left(0\right)\\ W_2& =& \underset{1}{\overset{4}{∫}}0dx\\ W_2& =& 0\end{array}$

(iii) From (4,0) to (4,2) , we have that x = 4

Then dx = 0

$\begin{array}{rcl}{W}_3& =& \underset{0}{\overset{2}{∫}}{\left(4\right)}^{2}y\left(0\right)-4{\left(y\right)}^{2}dy\\ W_3& =& \underset{0}{\overset{2}{∫}}-4y^{2}dy\\ W_3& =& \frac{-32}{3}\end{array}$

Find the total work done along the path.

$\begin{array}{rcl}W& =& W_1+W_2+W_3\\ W& =& \frac{1}{3}+0-\frac{32}{3}\\ W& =& \frac{-31}{3}\end{array}$

Hence the total work done is $\frac{86}{3}$ and work done along the given path is $\frac{-31}{3}$