Chapter 6: Applications of Definite Integrals - Section 6.4 - Areas of Surfaces of Revolution - Exercises 6.4 - Page 340: 10

$8 \pi \sqrt 5$

Integrate the integral to calculate the surface area as follows: We have: $S_{A}= (2 \pi)\int_{a}^{b} y \sqrt {1+(\dfrac{dy}{dx})^2}$ or, $ =(2 \pi)\int_{0}^{2}(2y) \cdot \sqrt {1+(2)^2} dy$ or, $=[4 \pi \sqrt 5y]_{0}^{2}$ Thus, $Surface \space Area=8 \pi \sqrt 5$