Line Integral Engineering Maths, Btech first year

Line Integral

Line Integral of a vector point function $\overrightarrow{F}(x,y,z)$ continuous along the curve 'C' from point A to B is defined as: $${\int }_C\overrightarrow{F}.d\overrightarrow{R}$$ $$\text{Here }\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$$ $$\text{So }d\overrightarrow{R}=dx\hat{i}+dy\hat{j}+dz\hat{k}$$ If the path of integration is a closed curve then line integral is: $$\oint \overrightarrow{F}.d\overrightarrow{R}$$

Irrotational vector

$\overrightarrow{F}$ is Irrotational if $\int \overrightarrow{F}.d\overrightarrow{R}=0$

Sample Problem 1

If $\overrightarrow{F}=(5xy-6x^2)\hat{i}+(2y-4x)\hat{j}$ evaluate ${\int }_C\overrightarrow{F}.d\overrightarrow{R}$ along the curve C in X-Y plane $y=x^3$ from point (1,1) to (2,8).

Solution

$$\overrightarrow{F}=(5xy-6x^2)\hat{i}+(2y-4x)\hat{j}$$ So now $\overrightarrow{F}.d\overrightarrow{R}$ $$={\int }_C[(5xy-6x^2)\hat{i}+(2y-4x)\hat{j}][dx\hat{i}+dy\hat{j}]$$ $$={\int }_C(5xy-6x^2)dx+(2y-4x)dy$$ Now the problem is, the integral that we have has terms of $x$ and $y$ mixed together i.e. $dx$ has $y$ variable and $dy$ has $y$ variable. To integrate we need them seperate. This is where we are going to use the equation of the curve given to us. The points given are used to identify the limits.
So there are three ways to integrate:
1) Reduce the terms with $dx$ to $x$ by replacing $y$ with $x$ and reduce terms with $dy$ to $y$ by replacing $x$ with $y$.
So using the given equation $y=x^3$ $$x=y^{1/3}$$ $$\Rightarrow {\int }_1^2[5x(x^3)-6x^2]dx+{\int }_1^8[2y-4(y{)}^{1/3}]dy$$ Note the limits: the integral with $dx$ terms has limits of the x-coordinates of the given points in the given order i.e. 1 to 2 and the integral with $dy$ terms has limits of the y-coordinates of the given points in the given order i.e. 1 to 8.

2) Reduce $dx$ and $x$ to $dy$ and $y$ respectively by using given equation. Here, $$\text{Put }x=y^{1/3}\text{ and }dx=\frac{1}{3}{y}^{-2/3}dy$$ $$\Rightarrow {\int }_1^8[5(y^{1/3})(y)-6(y^{1/3}{)}^2]\frac{1}{3}{y}^{-2/3}dy+[2y-4y^{1/3}]dy$$ Note that the limits are of y-coordinates only as we have reduced all the $x$ terms to $y$

3) Reduce $dy$ and $y$ to $dx$ and $x$ respectively by using given equation. Here, $$\text{Put }y=x^3\text{ and }dy=3x^{2}dx$$ $$\Rightarrow {\int }_1^2[5x(x^3)-6x^2]dx+[2x^3-4x](3x^{2}dx)$$ The limits hare are of x-coordinates only.
Here we will use the third way as integration is easier than the rest two methods. You can choose any of the three ways according to your convenience. $$\Rightarrow {\int }_1^2[5x(x^3)-6x^2]dx+[2x^3-4x](3x^{2}dx)$$ $$={\int }_1^2(5x^4-6x^2+6x^5-12x^3)dx$$ $$={[x^5-2x^3+x^6-3x^4]}_1^2$$ $$=32-16+64-48-(1-2+1-3)$$ $$=\boxed{{\displaystyle 35}}$$

Let's solve for a closed curve.

Sample Problem 2

Evaluate $${\int }_C(x^2+xy)dx+(x^2+y^2)dy$$ where C is a square formed by $x=\pm 1,y=\pm 1$.

Solution

First we trace the curve to identify the region using the given equations of curves $$x=\pm 1,y=\pm 1$$

Line Integral over the square ABCD is given by: $${\oint }_{ABCD}\overrightarrow{F}.d\overrightarrow{R}={\int }_{AB}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{BC}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{CD}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{DA}\overrightarrow{F}.d\overrightarrow{R}$$ Notice the naming used below the integral sign is in anticlockwise direction. That's the rule we will be using when writing limits for the curves i.e. in anticlockwise direction.
Now for Line Integral, we will integrate along each curve individually and then add them up. So along the curve AB :
The equation of the curve is $y=1$, so $$y=1\Rightarrow dy=0$$ Since here we will be integrating w.r.t. $dx$ the limits will be of x-coordinates of the points A(1,1) and B(-1,1) in anticlockwise direction $${\int }_{AB}\overrightarrow{F}.d\overrightarrow{R}={\int }_{AB}(x^2+xy)dx+(x^2+y^2)dy$$ $$={\int }_1^{-1}[x^2+x(1)]dx+0$$ $$={[\frac{x^3}{3}+\frac{x^2}{2}]}_1^{-1}$$ $$=(\frac{-1}{3}+\frac{1}{2})-(\frac{1}{3}+\frac{1}{2})$$ $$=\boxed{{\displaystyle \frac{-2}{3}}}$$ Along curve BC :
The equation of the curve is $x=-1$, so $$x=-1\Rightarrow dx=0$$ Since here we will be integrating w.r.t. $dy$ the limits will be of y-coordinates of the points B(-1,1) and C(-1,-1) in anticlockwise direction $${\int }_{BC}\overrightarrow{F}.d\overrightarrow{R}={\int }_{BC}(x^2+xy)dx+(x^2+y^2)dy$$ $$={\int }_1^{-1}0+[(-1{)}^2+y^2]dy$$ $$={\int }_1^{-1}(1+y^2)dy$$ $$={[y+\frac{y^3}{3}]}_1^{-1}$$ $$=(-1-\frac{1}{3})-(1+\frac{1}{2})$$ $$=\boxed{{\displaystyle \frac{-8}{3}}}$$ Along the curve CD The equation of the curve is $y=-1$, so $$y=-1\Rightarrow dy=0$$ Since here we will be integrating w.r.t. $dx$ the limits will be of x-coordinates of the points C(-1,-1) and D(1,-1) in anticlockwise direction $${\int }_{CD}\overrightarrow{F}.d\overrightarrow{R}={\int }_{CD}(x^2+xy)dx+(x^2+y^2)dy$$ $$={\int }_{-1}^1[x^2+x(-1)]dx+0$$ $$={\int }_{-1}^1[x^2-x]dx$$ $$={[\frac{x^3}{3}-\frac{x^2}{2}]}_{-1}^1$$ $$=\frac{1}{3}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}$$ $$=\boxed{{\displaystyle \frac{2}{3}}}$$ Along the curve DA :
The equation of the curve is $x=1$, so $$x=1\Rightarrow dx=0$$ Since here we will be integrating w.r.t. $dy$ the limits will be of y-coordinates of the points D(1,-1) and A(1,1) in anticlockwise direction $${\int }_{DA}\overrightarrow{F}.d\overrightarrow{R}={\int }_{DA}(x^2+xy)dx+(x^2+y^2)dy$$ $$={\int }_{-1}^{1}0+(1+y^2)dy$$ $$={[y+\frac{y^3}{3}]}_{-1}^1$$ $$=1+\frac{1}{3}+1+\frac{1}{3}$$ $$=\boxed{{\displaystyle \frac{8}{3}}}$$ Now adding together all the values: $${\oint }_{ABCD}\overrightarrow{F}.d\overrightarrow{R}={\int }_{AB}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{BC}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{CD}\overrightarrow{F}.d\overrightarrow{R}+{\int }_{DA}\overrightarrow{F}.d\overrightarrow{R}$$ $$=\left(\frac{-2}{3}\right)+\left(\frac{-8}{3}\right)+\frac{2}{3}+\frac{8}{3}$$ $$=0$$