Line Integrals - Mohsen Hamedi

The line integrals can be evaluated in two ways, depending on whether the curve $C$ is defined parametrically or by an explicit function. In either case the basic idea is to convert the line integral to a definite integral in a single variable.

Method of Evaluation – Curve Defined Parametrically

If $C$ is a smooth curve parameterized by $x=f(t),~ y=g(t),~ a \leq t \leq b$ , then we simply replace $x$ and $y$ in the integral by the functions $f(t)$ and $g(t)$ , and the appropriate differential $dx,~dy,$ or $ds$ by $f^\prime (t) dt ,~ g^\prime (t) dt$ , or $\sqrt{[f^\prime (t)]^2 + [g^\prime (t)]^2} dt$ . The expression $ds = \sqrt{[f^\prime (t)]^2 + [g^\prime (t)]^2} dt$ is called the differential of arc length. The integration is carried out with respect to the variable $t$ in the usual manner:

(1) $\begin{equation*}\int_C G(x,y) dx = \int_a^b G(f(t),g(t)) ~ f^\prime (t) dt,\end{equation*}$

(2) $\begin{equation*}\int_C G(x,y) dy = \int_a^b G(f(t),g(t)) ~ g^\prime (t) dt,\end{equation*}$

(3) $\begin{equation*}\int_C G(x,y) ds = \int_a^b G(f(t),g(t)) ~ \sqrt{[f^\prime (t)]^2 + [g^\prime (t)]^2} dt .\end{equation*}$

Method of Evaluation – Curve Defined by an Explicit Function

If the curve $C$ is defined by an explicit function $y=f(x), ~ a \leq x \leq b,$ we can use $x$ as a parameter. With $dy = f^\prime (x) dx$ and $ds = \sqrt{1+[f^\prime (x)]^2} dx$ , the foregoing line integrals become, in turn,

(4) $\begin{equation*}\int_C G(x,y) dx = \int_a^b G(x,f(x)) dx,\end{equation*}$

(5) $\begin{equation*}\int_C G(x,y) dy = \int_a^b G(x,f(x)) f^\prime (x) dx,\end{equation*}$

(6) $\begin{equation*}\int_C G(x,y) ds = \int_a^b G(x,f(x)) \sqrt{1+[f^\prime (x)]^2} dx.\end{equation*}$

YouTube Videos

Watch these two awesome videos on line integral and its physical interpretation.

Reference

Dennis G. Zill. Advanced Engineering Mathematics, $6^{th}$ edition. Jones $&$ Bartlett Learning. 2016.