Question

Evaluate the line Integral f (x²+xy)dx + (x² + y2)dy where C is the square formed by lines x = +1, y = +1

Answer 1

C is the square formed by lines.

 x = ± 1 & y = ±1.

\(∮_c\)[(x²+xy)dx + (x²+y²)dy]

= (\(\int_{AB}\) + \(\int_{BC}\) + \(\int_{CD}\) + \(\int_{DA}\)) [(x²+xy)dx + (x²+y²)dy]

On line AB,

x = 1 & y is variable

∴ dx = 0 & -1 ≤ y ≤ 1

On line BC,

y = 1 & x is variable

∴ dx = 0 & 1 ≤ x ≤ -1

On line CD,

x = -1 & y is variable

∴ dx = 0 & 1 ≤ y ≤ -1

On line DA,

y = -1 & x is variable

∴ dy = 0 & -1 ≤ x ≤ -1

= \(\int_{-1}^1(1+y^2)dy\) + \(\int_{1}^{-1}(x^2+x)dx\) + \(\int_{1}^{-1}(1+y^2)dy\) + \(\int_{-1}^1(x^2-x)dx\)

= \([y + \frac{y^3}{3}]_{-1}^1 ​\) + \([\frac{x^3}{3}\) + \(\frac{x^2}{2}]^{-1}_1\) + \([y + \frac{y^3}{3}]_1^{-1}\)  + \([\frac{x^3}{3}\)-\(\frac{x^2}{2}]^1_{-1}\)

= (1 + \(\frac{1}{3}\)) -  (-1 - \(\frac{1}{3}\)) + (\(\frac{-1}{3}\) + \(\frac{1}{2}\)) - (\(\frac{1}{3}\)+\(\frac{1}{2}\)) + (-1-\(\frac{1}{3}\)) - (1+\(\frac{1}{3}\)) + (\(\frac{1}{3}\)-\(\frac{1}{2}\)) - (\(\frac{-1}{3}\) - \(\frac{1}{2}\))

= \(\frac{-2}{3}\) + 0 + \(\frac{2}{3}\) - 0

= 0