The concept of line integrals, also path integrals or curvilinear integrals extend the idea of ** definite integrals**. We have done the definite integral in two dimensional

**plane to find the area of curve**

*X-Y**f(x)*between two points,

*x=a*, and

*x=b*. $$\int_{x=a}^{b}{f(x)}\;dx$$

Here, **f(x****)** is the height of the Area, * dx* is the infinitesimal

__of infinitely small area__

*base***, where,**

*f(x)dx**dx=Δx*such that

*Δx→0*. Hence we sum these

__infinite numbers__of infinitesimally small rectangles between

**and**

*a***, that gives the total area of the curve between these limits. However line integrals based on the same idea but it extend from two dimensional to three dimensional curve.**

*b*## Line Integrals

Now let say some *parametric equations**x=g(t)*, *y=h(t) *create a ** Path** on

**plane for arbitrary**

*X-Y***within**

*t***. This curve is now similar to our previous two-dimensional curve. But if we have another function**

*a≤ t ≤b**that associates with every point to the*

**f(x,y)****plane with some values. Then our curve in three dimensional space will look like this:**

*X-Y*For every value of *x*, and *y*,* f(x,y)* has a value along ** Z**-axis. So in line integrals, we want to find the area of the “curvy wall or curtain” that goes along the Path. Imagine,

*f(x, y)*is the roof over

*X-Y*plane. then we want to know the area of the curve (

*wall*) along X-Y plane under the roof such that the wall touches the roof at

*f(x,y)*.

Let ** dS** is the little change in distance of the wall, and

*f(x,y)*is the height of the wall at that point. So, if we multiply them, then we will find the area of that little rectangle,

*f(x,y)dS*. Here

**refers to as infinitesimal change in**

*dS***of the curve with respect to**

*arc length***. So,**

*t***is the infinitely narrow width, therefore, there are infinite number of**

*dS***along the path. So if we take the sum of all infinitely small rectangles from**

*dS***, to**

*t=a***, then this will give us the area of wall:**

*t=b*$$\int_{t=a}^{t=b}{f(x, y)}\;dS$$

This is the same idea of definite integral on *X-Y* plane. But, ** dS** actually indicates the infinitesimally small changes in X and Y direction. Therefore, we can find

**using pythagorean theorem. So, $dS=\sqrt{dx^{2}+dy^{2}}$. therefore our line integrals become:**

*dS*$$\int_{t=a}^{t=b}{f(x, y)}\;\sqrt{dx^{2}+dy^{2}}$$

As *f* is a function of *x, y*. *x* is a function of *t*, also *y* is a function of *t*. Therefore, we can rewrite the our line integral in following form:

$$\int_{t=a}^{t=b}{f(x(t), y(t))}\;\sqrt{dx^{2}+dy^{2}}$$

Now if we multiply the x, y term with $\frac{dt}{dt}$, then we get,

$$\frac{dt}{dt}\sqrt{dx^{2}+dy^{2}}\\

= \sqrt{\frac{1}{dt^{2}}(dx^{2}+dy^{2})}. dt\\

\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}. dt$$

As we know ** x** is a function of

**,**

*t***, therefore $\frac{dx}{dt}=g'(t)$. Also**

*x=g(t)***is a function of**

*y***,**

*t***, therefore, $\frac{dy}{dt}=h'(t)$. So our path integral finally gets the form:**

*y=h(t)*$$\bbox[5px,border:2px solid black] {\int_{t=a}^{t=b}{f(x(t), y(t))}\;\sqrt{(g'(t))^{2}+(h'(t))^{2}}dt}$$

Here in line integrals, what we are essentially doing is, we are taking integral over a curve, or over a line in an interval of ** t=a** to

**, on the**

*t=b***. similar to definite integral, here, $dS=\sqrt{(g'(t))^{2}+(h'(t))^{2}}$ is just the infinitesimal change in the arc length with respect to some parameter**

*x-axis***. and**

*t***is the height at that point.**

*f(x(t), y(t))*//This is the general intuition on Line integrals | Path or Curvilinear Integrals. Next we will see some problem solving on it. Read more articles on ** vector analysis** and

**.**

*coordinate geometry*