Line integrals | Path or Curvilinear Integrals | Vector Analysis

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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 X-Y plane to find the area of curve 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 base of infinitely small area f(x)dx, where, dx=Δx such that Δx→0. Hence we sum these infinite numbers of infinitesimally small rectangles between a and b, 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.

Line Integrals | Path Or Curvilinear Integrals | Vector Analysis
Figure: Definite Integral

Line Integrals

Now let say some parametric equations x=g(t), y=h(t) create a Path on X-Y plane for arbitrary t within a≤ t ≤b. This curve is now similar to our previous two-dimensional curve. But if we have another function f(x,y) that associates with every point to the X-Y plane with some values. Then our curve in three dimensional space will look like this:

Line Integrals | Path Or Curvilinear Integrals | Vector Analysis - CS School
Figure: Curve in 3d space

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 dS refers to as infinitesimal change in arc length of the curve with respect to t. So, dS 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 t=a, to t=b,  then this will give us the area of wall:

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

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

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

As f is a function of x, yx 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, x=g(t), therefore $\frac{dx}{dt}=g'(t)$. Also y is a function of t, y=h(t), therefore, $\frac{dy}{dt}=h'(t)$. So our path integral finally gets the form:

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

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 t=b, on the x-axis. 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 t. and f(x(t), y(t)) is the height at that point.

//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.

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