Parameterize C by the two vector functions,
r(t) = (1 - t) (0, 0) + t (6, 0) = (6t, 0)
with 0 ≤ t ≤ 1, and
s(t) = (6 cos(t), 6 sin(t))
with 0 ≤ t ≤ π/2.
Then the line integral over C is equal to the sum of the line integrals over each path:
[tex]\displaystyle \int_C dS = \int_0^1 \|r'(t)\| \, dt + \int_0^{\frac{\pi}2} \|s'(t)\| \, dt[/tex]
[tex]\displaystyle \int_C dS = \int_0^1 \sqrt{6^2 + 0^2} \, dt + \int_0^{\frac{\pi}2} \sqrt{(-6\sin(t))^2 + (6\cos(t))^2} \, dt[/tex]
[tex]\displaystyle \int_C dS = 6 \int_0^1 dt + 6 \int_0^{\frac{\pi}2} dt[/tex]
[tex]\displaystyle \int_C dS = 6 + \frac{6\pi}2 = \boxed{6+3\pi}[/tex]
