Answer
$$ \frac{29\pi }{30}$$
Work Step by Step
$y= x^2,\ \ x=y^2 ; \quad$ about the $x=-1$
First, we find the intersection points
\begin{aligned} x^4&=x\\
x^4-x&=0\\
x(x^3-1)&=0 \end{aligned}
Then
\begin{aligned} x=0,\ \ x=1 , \to \ \ (0,0), (1,1)\end{aligned}
Here the oute radius is $\sqrt{y} +1 $ and the inner is $ y^2+1$
\begin{aligned}
A(y)&= \pi [f^2(y)-g^2(y)]\\
&= \pi[ (\sqrt{y}+1)^2- (y^2+1)^2]\\
&= \pi [-y^4-2y^2+y+2\sqrt{y}]
\end{aligned}
Then the volume of the solid given by
\begin{aligned}
V&= \int_a^bA(y)dy\\
&= \pi\int_0^1 (-y^4-2y^2+y+2\sqrt{y})dy\\
&=\pi (-\frac{y^5}{5}-\frac{2y^3}{3}+\frac{y^2}{2}+\frac{4}{3}y^{\frac{3}{2}})\bigg|_0^1\\
&= \frac{29\pi }{30}
\end{aligned}