Finding the volume of the area between two curves when rotated about the y-axis I keep going down a rabbit hole when answering this question:

Find the volume of the area lying in the first quadrant and bounded by the $y$-axis, the curve $y = x^3$ and the line $y = 3x + 2$, when it performs one revolution about the $y$-axis.

We need to find the ordinates first:
$$\begin{align}
x^3 & = 3x + 2 \\
0 & = x^3 - 3x + 2 \\
0 & = (x+1)^2(x-2)
\end{align}$$
This gives us the co-ordinates where both meet: $(-1, -1)$ and $(2, 8)$. But since we're only interested in the first quadrant, we'll use $y = 8$ and $y = 0$ as the ordinates to bound our area.
Let $A_1$ be the area between the $y$-axis and $y=x^3$, and let $A_2$ be the area between the $y$-axis and $y = 3x + 2$. The total volume, $V_T$, will be found by $V_1 - V_2$; where $V_1$ is the volumes found by revolving $A_1$, and $V_2$ is the volume found by revolving $A_2$.
$$\begin{align}
A_1 & = \pi(x^2) =\pi\left(\sqrt[3]{y}\right)^2=\pi\left(y^{\frac{2}{3}}\right) \\ 
A_2 & = \pi(x^2) = \pi\left[\left(\frac{y-2}{3}\right)^2\right] = \pi\left[\frac{y^2}{9} - \frac{4y}{9}+\frac{4}{9}\right]
\end{align}$$
Now using the volume of revolution equation:
$$\begin{align}
V_1 & = \pi \int _{8} ^{0} (y^{\frac{2}{3}})dy = \pi\left[\frac{3y^{\frac{5}{3}}}{5}\right]_0^8\\
& = \frac{3}{5}(32)\pi = \frac{96}{5}\pi\\
\\
V_2 & = \pi \int _{8} ^{0} \left(\frac{y^2}{9} - \frac{4y}{9}+\frac{4}{9}\right)dy = \pi\left[\frac{y^3}{27} - \frac{2y^2}{9}+\frac{4y}{9}\right]_0^8\\
& = \left( \frac{512}{27} - \frac{128}{9} + \frac{32}{9} \right)\pi = \frac{227}{27}\pi\\
\\
V_T & = V_1 - V_2 = \frac{1457}{135}\pi
\end{align}$$ 
But the expected answer is $\frac{56}{5}\pi$. Where am I going wrong?

When calculating $V_2$, you need to integrate from $2$ to $8$ because for $0 \le y < 2$, we have $x=(y-2)/3<0$, which is left of the $y$-axis (plot the graph to see this more clearly).
