Dated: 30-10-2024

Examples

Example

Evaluate the following along the boundaries of a rectangle with vertices \(A(1, 0)\), \(B(3, 0)\), \(C(3, 2)\) and \(D(1, 2)\).

\[\oint (xy \, dx + (1 + y^2)dy)\]

\[I_c = \oint (xy \, dx + (1 + y^2)dy)\]

We can divide \(c\) into 4 paths, representing the sides of the rectangle.

\[\overline{AB} : c_1 : y = 0 \text { from } x = 1 \text{ to } 3\]

Putting this into our integrand,¹ we get

\[I_{c_1} = 0\]

\[\overline{BC} : c_2 : x = 3 \therefore dx = 0\]

Putting this into our integrand,¹ we get

\[I_{c_2}=\int_{0}^{2}(1+y^{2})dy=\left[y+\frac{y^{3}}{3}\right]_{0}^{2}\]

\[= \frac {15} 3\]

\[c_3 : \overline{CD} : y = 2 \therefore dy = 0\]

\[I_{c_3}=\int_{3}^{1}2xdx=\left[x^{2}\right]_{3}^{1}=-8\]

\[c_4 : \overline{DA} : x = 1 \therefore dx = 0\]

\[I_{c_4}=\int_{2}^{0}(1+y^{2})dy=\left[y+\frac{y^{3}}{3}\right]_{2}^{0}\]

\[= - \frac {15} 3\]

Therefore

\[I_c = I_{c_1} + I_{c_2} + I_{c_3} + I_{c_4} = -8\]

\[\because I = \int_C F_t ds\]

We can relate a function² \(f(x, y)\) which depends on the position of the point.

\[\because I = \int_C f(x, y) ds\]

which we can of course convert into an integral¹ depending upon

\[I = \int_C f(x, y) \frac {ds}{dx} dx\]

Where

\[\frac{ds}{dx} = \sqrt{\left(\frac{dx}{dx}\right)^2 + \left(\frac{dy}{dx}\right)^2}\]

\[\int_{C}f(x,y)dx=\int_{x_{1}}^{x_{2}}f(x,y)\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\]

If the integrant¹ seen to be an exact differential³ then the line integral³ is independent of the path taken.