Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 32

$\dfrac{ \pi}{2}$

Area $=\pi r^2=\pi (\dfrac{\sqrt {2y}}{y^2+1})^2=\dfrac{2 \pi y}{(y^2+1)^2}$ We integrate the integral to calculate the volume as follows: $V= \times \int_{0}^{1} \dfrac{2 \pi y}{(y^2+1)^2} dy$ Suppose $u =t^2+1 \implies dt=2y dy$ Now, $V= \pi \in_{1}^2 \dfrac{dt}{t^2}$ or, $= \pi (-\dfrac{1}{t})_{1}^{2}=\dfrac{ \pi}{2}$