Answer
$\text{The volume is}$
\begin{align}
V =\frac{\pi}{2}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
x = y^2 \ and \ x= y
\end{align}
$\text{The function revolves around y = -1. Thus,}$
\begin{align}
y = \sqrt x
\end{align}
$\text{The intersections of two functions are}$
\begin{align}
\sqrt x = x \Rrightarrow x = 0 \ and \ x = 1 \Rrightarrow y = 0 \ and \ y = 1
\end{align}
$\text{The volume is}$
\begin{align}
& V =\pi \int_0^1 \left((\sqrt x + 1)^2 - (x+1)^2\right) \ dx = \pi \int_0^1 \left(-x^2-x+2\sqrt x \right) \ dx = \\ & = \pi \left(-\frac{1}{3} - \frac{1}{2} + \frac{4}{3}\right) = \frac{\pi}{2}
\end{align}
