Answer:
Step-by-step explanation:
Given that
[tex]r(t) = (6cost, 6sint, t), 0\leq t\leq 2\pi\\r'(t) = (-6sint, 6cost, 1),\\||r'(t)||=\sqrt{(-6sint)^2 +(6cost)^2+1} =\sqrt{37}[/tex]
Hence arc length = [tex]\int\limits^a_b {||r'(t)||} \, dt[/tex]
Here a = 0 b = 2pi and r'(t) = sqrt 37
Hence integrate to get
[tex]\int\limits^{2\pi} _0 {\sqrt{37} } \, dt\\ =\sqrt{37} (t)\\=2\pi\sqrt{37}[/tex]
