Volume generated by rotating around y-axis, curve $y=x^3$ and the lines $y=0$ and $x=2$ 
Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$

I first found what $x=2$ would be in terms of y.
$$y= (2)^3
\\ = 8$$
And in terms of y, the original equation becomes:
$$y=x^3
\\ x= y^{1 \over 3}$$
So, 
$$V= \int^8_0 (y^{1 \over 3})^2 \pi dy
\\ = \pi \int_0^8 y^{2\over 3}dy
\\ =\pi \bigg[ y^{5\over 3} ({3 \over 5}) \bigg]^8_0
\\ = (8^{5 \over 3})({3 \over 5})\pi
\\ = {96 \over 5} \pi$$ 
Therefore the answer is ${96 \over 5} \pi$ units $^3$
However, the answer is supposed to be ${64 \over 5} \pi$ units${^3}$
What went wrong?

Washer method is quite easy. Express the washer area including its hole and extrude it along $y$.
$$
V=\pi\int_{0}^{8}\left(2^2-x^2\right)\,dy=\pi\int_{0}^{2}\left(2^2-x^2\right)\,3x^2 dx= $$
$$
V=3\pi\int_{0}^{2}\left(4-x^2\right)\, \cdot x^2 \, dx = ... = \frac{64 \pi}{5}$$
