91Ó°ÊÓ

$\mathbf{F}(x,y)=x^3{y}^2\mathbf{i}+(y-x)\mathbf{j}$

The work done by a vector field F moving a particle along the curve C is computed by the following line integral:

${∫}_C\mathbf{F}·d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${∫}_C\mathbf{F}·d\mathbf{r}$to evaluate the required work done.

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)âŸ\copyright$for $a≤t≤b$.

If a vector field is defined on R, then,

${∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA·"…..\left(1\right)$

For the vector field $\mathbf{F}(x,y)=x^3{y}^2\mathbf{i}+(y-x)\mathbf{j}$,

$F_1(x,y)=x^3{y}^2$and $F_2(x,y)=y-x.$

Now, first, find $\frac{∂F_2}{∂x}$and $\frac{∂F_1}{∂y}$.

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}(y-x) \\ =-1, \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(x^3{y}^2\right) \\ =2x^{3}y \end{gathered}$$

Now, use Green's Theorem (1) to evaluate the integral ${∫}_C\mathbf{F}·d\mathbf{r}$as follows:

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ ={âˆ\neg }_R\left(-1-2x^{3}y\right)dA \\ =-{âˆ\neg }_R\left(1+2x^{3}y\right)dA……\left(2\right) \end{gathered}$$

Here, the boundary curve C is a triangle with vertices $(1,1),(2,2),(3,1)$, starting and ending at $(2,2)$.

So the region R bounded by this triangle is shown in the following figure.

Viewed it as a x-simple region, then the region of integration will be,

$R=\left\{\right(x,y)âˆ\pounds y≤x≤4-y,1≤y≤2\}.$

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}=-{∫}_R\left(1+2x^{3}y\right)dA \\ =-{∫}_1^2{∫}_y^{4-y}\left(1+2x^{3}y\right)dxdy \\ =-{∫}_1^2\left[{∫}_y^{4-y}\left(1+2x^{3}y\right)dx\right]dy \\ =-{∫}_1^2{\left[x+\frac{1}{2}{x}^{4}y\right]}_y^{4-y}dy \\ =-{∫}_1^2\left[\left((4-y)+\frac{1}{2}(4-y{)}^{4}y\right)-\left(y+\frac{1}{2}{y}^{4}y\right)\right]dy \\ =-{∫}_1^2\left[4-y+\frac{1}{2}(4-y{)}^{4}y-y-\frac{1}{2}{y}^5\right]dy \\ =-{∫}_1^2\left[4-2y-\frac{1}{2}{y}^5+\frac{1}{2}(4-y{)}^{4}y\right]dy \\ =-{\left[4y+63y^2-\frac{128}{3}{y}^3+12y^4-\frac{8}{5}{y}^5\right]}_1^2 \\ =-\left[\left(4·2+63·2^2-\frac{128}{3}·2^3+12·2^4-\frac{8}{5}·2^5\right)\right. \\ \left.-\left(4·1+63·1^2-\frac{128}{3}·1^3+12·1^4-\frac{8}{5}·1^5\right)\right] \\ =-\left(\frac{371}{15}\right) \\ =-\frac{371}{15}. \end{gathered}$$

Therefore, the required work done is $-\frac{371}{15}$.