Chapter 6: Applications of Definite Integrals - Section 6.4 - Areas of Surfaces of Revolution - Exercises 6.4 - Page 340: 15

$2 \pi$

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_{1/2}^{3/2} \sqrt {2x-x^2} \sqrt {\dfrac{1}{2x-x^2}} dx$ or, $=(2 \pi) \int_{1/2}^{3/2} x dx$ or, $ = 3 \pi -\pi $ Thus, $Surface \space Area=2 \pi$