Answer
$\int_Cf(x,y) ds = 16\pi$
Work Step by Step
The path is given by $C:x^2+y^2 = 4$
This can be written as , $\textbf r(t) = x(t) \textbf i + y(t) \textbf j = 2\cos t \textbf i + 2\sin t \textbf j, 0 \le t \le 2\pi$
$\therefore x(t) = 2\cos t , y(t) = 2\sin t$
$\textbf r' = -2\sin t \textbf i +2\cos t \textbf j, \vert \textbf r' \vert = 2$
$f(x,y) = x^2-y^2+4 = 4\cos ^2t - 4\sin^2 t +4 = 4\cos 2t + 4 \\
\therefore \int_C f(x,y) ds = \int f(x,y) \vert \textbf r' \vert dt\\
= \int_{t=0}^{2\pi} (4\cos 2t +4).2.dt = 8[\frac{\sin 2t}{2} + t]_0^{2\pi} = 16\pi$
