Chapter 6: Applications of Definite Integrals - Section 6.4 - Areas of Surfaces of Revolution - Exercises 6.4 - Page 340: 9

$4 \pi \sqrt 5$

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}^{4}(\dfrac{x}{2}) \cdot \sqrt {1+\dfrac{1}{4}} dx$ or, $=[\dfrac{\sqrt 5 \pi x^2}{4} ]_0^4$ Thus, $Surface \space Area=4 \pi \sqrt 5$