The area:
[tex]A = 2 \pi \lim_{b\to \infty} ( \int\limits^\infty_0 {e ^{-6x} } \, dx) [/tex]
Substitution: u = - 6 x
du = - 6 dx
dx = -1/6 du
∫ -1/6 e^(u) du = - 1/6 e^(u) = - 1/6 e^(-6x)
[tex]A = -1/6 * 2 \pi \lim_{b \to \infty} (\int\limits^\infty_0 \, e ^{-6b} db) = \\ =- \pi /3 * ( e ^{-\infty} -e ^{0}) = \\ = - \pi /3 * ( - 1 ) = [/tex]
= π / 3
