Chapter 3, Problem 40P

(a)

To determine

The value of the line integral ${\displaystyle {\oint }_{L}F\cdot dl}$.

Explanation of Solution

Given:

The vector $\left(F\right)$ is $x^{2}y{a}_x-y{a}_y$.

The boundaries for $L$ are as shown in the figure-(1).

Figure-(1)

Calculation:

The line distribution of the given figure is as shown in figure (2).

Figure-(2)

From the figure (2), the equations for lines $1$, $2$, and $3$ are as follows.

For line $1$,

  $x=y$

  $dx=dy$

The limit of $\left(x\right)$ is $0<x<1$.

Calculate the value of line integral $\left({\displaystyle {\oint }_{1}F\cdot dl}\right)$ using the relation.

  ${\displaystyle {\oint }_{1}F\cdot dl}={\displaystyle {\oint }_{1}F\cdot \left(dx{a}_x+dy{a}_y\right)}$

  $\begin{array}{c}{\displaystyle {\oint }_{1}F\cdot dl}={\displaystyle {\oint }_1\left(x^{2}y{a}_x-y{a}_y\right)\cdot \left(dx{a}_x+dy{a}_y\right)}\\ ={\displaystyle {\oint }_1{x}^{2}ydx-ydy}\\ ={\displaystyle {\int }_0^1{x}^2\left(x\right)dx-\left(x\right)\left(dx\right)}\\ ={\displaystyle {\int }_0^1\left(x^3-x\right)dx}\end{array}$

  $\begin{array}{c}{\displaystyle {\oint }_{1}F\cdot dl}={\left[\frac{x^4}{4}-\frac{x^2}{2}\right]}_0^1\\ =\left(\frac{1^4}{4}-\frac{1^2}{2}\right)-\left(\frac{0^4}{4}-\frac{0^2}{2}\right)\\ =-\frac{1}{4}\end{array}$

For line $2$,

  $y=-x+2$

  $dy=-dx$

The limit of $\left(x\right)$ is $1<x<2$.

Calculate the value of line integral $\left({\displaystyle {\oint }_{2}F\cdot dl}\right)$ using the relation.

  ${\displaystyle {\oint }_{2}F\cdot dl}={\displaystyle {\oint }_{2}F\cdot \left(dx{a}_x+dy{a}_y\right)}$

  $\begin{array}{c}{\displaystyle {\oint }_{2}F\cdot dl}={\displaystyle {\oint }_2\left(x^{2}y{a}_x-y{a}_y\right)\cdot \left(dx{a}_x+dy{a}_y\right)}\\ ={\displaystyle {\oint }_2{x}^{2}ydx-ydy}\\ ={\displaystyle {\int }_1^2{x}^2\left(-x+2\right)dx-\left(-x+2\right)\left(-dx\right)}\\ ={\displaystyle {\int }_1^2\left(-x^3+2x^2-x+2\right)dx}\end{array}$

  $\begin{array}{c}{\displaystyle {\oint }_{2}F\cdot dl}={\left[-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{x^2}{2}+2x\right]}_1^2\\ =\left(-\frac{2^4}{4}+\frac{2{\left(2\right)}^3}{3}-\frac{2^2}{2}+2\left(2\right)\right)-\left(-\frac{1^4}{4}+\frac{2{\left(1\right)}^3}{3}-\frac{1^2}{2}+2\left(1\right)\right)\\ =\left(\frac{10}{3}\right)-\left(\frac{23}{12}\right)\\ =\frac{17}{12}\end{array}$

For line $3$,

  $y=0$

  $dy=0$

The limit of $\left(x\right)$ is $0<x<2$.

Calculate the value of line integral $\left({\displaystyle {\oint }_{3}F\cdot dl}\right)$ using the relation.

  ${\displaystyle {\oint }_{3}F\cdot dl}={\displaystyle {\oint }_{3}F\cdot \left(dx{a}_x+dy{a}_y\right)}$

  $\begin{array}{c}{\displaystyle {\oint }_{3}F\cdot dl}={\displaystyle {\oint }_3\left(x^{2}y{a}_x-y{a}_y\right)\cdot \left(dx{a}_x+dy{a}_y\right)}\\ ={\displaystyle {\oint }_3{x}^{2}ydx-ydy}\\ =-{\displaystyle {\int }_0^2{x}^2\left(0\right)dx-\left(0\right)\left(0\right)}\\ =-{\displaystyle {\int }_0^2\left(0\right)dx}\end{array}$

  ${\displaystyle {\oint }_{3}F\cdot dl}=0$

Calculate the value of total line integral $\left({\displaystyle {\oint }_{L}F\cdot dl}\right)$ using the relation.

  ${\displaystyle {\oint }_{L}F\cdot dl}={\displaystyle {\oint }_{1}F\cdot dl}+{\displaystyle {\oint }_{2}F\cdot dl}+{\displaystyle {\oint }_{3}F\cdot dl}$

  $\begin{array}{c}{\displaystyle {\oint }_{L}F\cdot dl}=-\frac{1}{4}+\frac{17}{12}+0\\ =\frac{14}{12}\\ =1.1667\end{array}$

Thus, the value of line integral ${\displaystyle {\oint }_{L}F\cdot dl}$ is $\underset{\_}{1.1667}$.

(b)

To determine

The value of surface integral ${\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}$.

Explanation of Solution

Given:

The vector $\left(F\right)$ is $x^{2}y{a}_x-y{a}_y$.

The area $\left(S\right)$ for boundary $\left(L\right)$ is as shown in the figure-(3).

Figure-(3)

Calculation:

Calculate the curl of vector $\left(\nabla \times F\right)$ using the relation.

  $\nabla \times F=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right|$

  $\begin{array}{c}\nabla \times F=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^{2}y& \left(-y\right)& 0\end{array}\right|\\ =a_x\left(\frac{\partial }{\partial y}\left(0\right)-\frac{\partial }{\partial z}\left(-y\right)\right)-a_y\left(\frac{\partial }{\partial x}\left(0\right)-\frac{\partial }{\partial z}{x}^{2}y\right)+a_z\left(\frac{\partial }{\partial x}\left(-y\right)-\frac{\partial }{\partial y}{x}^{2}y\right)\\ =a_x\left(0-0\right)-a_y\left(0-0\right)+a_z\left(0-x^2\right)\\ =-x^2{a}_z\end{array}$

The line distribution of the given figure is as shown in figure-(4).

Figure-(4)

From the figure (2), the equations for line $1$ and $2$ are as follow.

For line $1$,

  $x=y$

The limit of $\left(x\right)$ is $0<x<1$.

For line $2$,

  $y=-x+2$

The limit of $\left(x\right)$ is $1<x<2$.

Calculate the value of surface integral $\left({\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}\right)$ using the relation.

  ${\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}={\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot \left(dydz{a}_x+dxdz{a}_y-dxdy{a}_z\right)}$

  $\begin{array}{c}{\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}={\displaystyle {\oint }_S\left(-x^2{a}_z\right)\cdot \left(dydz{a}_x+dxdz{a}_y-dxdy{a}_z\right)}\\ ={\displaystyle {\oint }_S{x}^{2}dxdy}\\ ={\displaystyle {\int }_0^1{\displaystyle {\int }_0^x{x}^{2}dxdy}}+{\displaystyle {\int }_1^2{\displaystyle {\int }_0^{-x+2}{x}^{2}dxdy}}\\ ={\displaystyle {\int }_0^1{x}^{2}dx{\left[y\right]}_0^x}+{\displaystyle {\int }_1^2{x}^{2}dx{\left[y\right]}_0^{-x+2}}\end{array}$

  $\begin{array}{c}{\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}={\displaystyle {\int }_0^1{x}^2\left(x-0\right)dx}+{\displaystyle {\int }_1^2{x}^2\left(-x+2-0\right)dx}\\ ={\displaystyle {\int }_0^1{x}^{3}dx}+{\displaystyle {\int }_1^2\left(-x^3+2x^2\right)dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1+{\left[-\frac{x^4}{4}+\frac{2x^3}{3}\right]}_1^2\\ =\left(\frac{1^4}{4}-\frac{0^4}{4}\right)+\left[\left(-\frac{2^4}{4}+\frac{2{\left(2\right)}^3}{3}\right)-\left(-\frac{1^4}{4}+\frac{2{\left(1\right)}^3}{3}\right)\right]\end{array}$

  $\begin{array}{c}{\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}=\frac{1}{4}+\left(-4+\frac{16}{3}+\frac{1}{4}-\frac{2}{3}\right)\\ =\frac{1}{4}+\frac{11}{12}\\ =\frac{14}{12}\\ =1.667\end{array}$

Thus, the value of surface integral ${\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}$ is $\underset{\_}{1.1667}$.

(c)

To determine

Whether the Stokes’s theorem is satisfied or not.

Explanation of Solution

The value of line integral ${\displaystyle {\oint }_{L}F\cdot dl}$ is $1.667$.

The value of surface integral ${\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}$ is $1.667$.

The Stokes’s theorem states that if $F$ is a continuously differentiable vector point defined on an open surface bounded by the curve $L$ in a positive direction then,

  ${\displaystyle {\oint }_{L}F\cdot dl}={\displaystyle {\oint }_S\left(\nabla \times F\right)\cdot dS}$

From the above calculations in subpart (a) and subpart (b), these values are same.

Thus, the Stokes’s theorem is satisfied.