$y=x^2, y=1$; about the line $y=-1$

Alright so I have the sketch drawn but I cannot figure out if I'm doing this correctly because the $y=-1$ is throwing me off. The answer I got is $4/5\pi$.

I got that answer by using horizontal slices $R=\sqrt{y}=x$ and $r=y$. 
$$
\begin{align}
V&=2\pi \int\_0^1 y(\sqrt{y})= \int\_0^1(y)^{3/2}\\
&=2\pi \frac{2}{5}y^{5/2} \bigg|\_0^1=\frac{4}{5}\pi
\end{align}
$$
Help if I went about this the wrong way. This stuff is just completely confusing.