Dated: 30-10-2024
Line Integral
Example
Evaluate \(\int_C (x + 3y) dx\) from \(A(0, 1)\) and \(B(2, 5)\) along the curve \(y = 1 + x^2\)
Solution
\[I = \int_C (P dx + Q dy)\]
\[\because Q = 0\]
\[\because C: y = 1 + x^2\]
\[\therefore I = \int_0^2 (x + 3(1 + x^2)) dx\]
\[= \int_0^2 (x + 3 + 3x^2)dx\]
\[=\left[\frac{x^{2}}{2}+3x+x^{3}\right]_{0}^{2}=16\]
Example
Evaluate the following from \(O(0, 0)\) to \(A(1, 0)\) along the line1 \(y = 0\) and then from \(A(1, 0)\) to \(B(1, 4)\) along the line1 \(x = 1\).
\[I=\int_{C}((x^{2}+2y)dx+xydy)\]
Solution
\[I_C = I_{c_1} + I_{c_2}\]
\[C = \overline{OA} + \overline{AB}\]
\[c_1: y = 0 \therefore dy = 0\]
\[c_2: x = 1 \therefore dx = 0\]
Substituting these values in our integral,2 we get
\[I_{c_1}=\int_{0}^{1}x^{2}dx=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}\]
Similarly,
\[I_{c_2}=\int_{0}^{4}ydy=\left[\frac{y^{2}}{2}\right]_{0}^{4}=8\]
\[\therefore I = 8 + \frac 1 3 = \frac {25} 3\]
Properties of line Integrals
-
\[\int_{C}Fds=\int_{C}\{Pdx+Qdy\}\]
-
\[\int_{AB}Fds=-\int_{BA}Fds\]
-
We can have paths which are parallel to the axes.
- For being parallel to \(y\) axis, \(\int_C P dx = 0\) and \(I_c = \int_c Q dy\)
- For being parallel to \(x\) axis, \(\int_C Q dy = 0\) and \(I_c = \int_c P dx\)
- We can divide a path into sub paths. Such as \(I_{AB} = I_{AK} + I_{KB}\)
References
Read more about notations and symbols.