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?
 A: The shell method would be easiest to use here:
$$
V=2\pi\int_{0}^{2}x\cdot x^3\,dx=2\pi\int_{0}^{2}x^4\,dx=
2\pi\frac{x^5}{5}\bigg|_{0}^{2}=\\
\frac{2\pi}{5}\left(2^5-0^5\right)=
\frac{2\pi}{5}\cdot 32=\frac{64\pi}{5}\ \text{cubic units}.
$$
Or you can use the washer method which is going to be a bit messier. In that case, you're subtracting the volume of a cylinder of radius $2$ and height $8$ from the volume you get by revolving the curve $x=\sqrt[3]{y}$ around the $y$-axis:
$$
V=\int_{0}^{8}\left(\pi\cdot 2^2-\pi\left(\sqrt[3]{y}\right)^2\right)\,dy=
\pi\int_{0}^{8}\left(4-y^{\frac{2}{3}}\right)\,dy=\frac{64\pi}{5}\ \text{cubic units}.
$$
Wolfram Alpha check.
A: The volume required is between the lines y=0, x=2 and curve y=x^3. 
If you visualize it, the required volume is :
V1-V2 
where V1 = the volume of cylinder formed by line y=2 and x=2. 
V2 = the volume enclosed by the curve y=x^3 around y axis. 
V1 = pi*r^2*h.
r=2, h = 8.
so V1 = 4*8*pi = 32 pi
V2 = 96/5 pi
V1-V2 = 32pi - 96/5pi = 64/5 pi
Please pardon me as I dont know the mathML.
A: You are calculating the empty volume between the rotated function and the y-axis. This is because for every y-value, you are summing the distance between the y-axis and the function y=x^3, rather than the distance between the function y=x^3 and your boundary of x=2. As mentioned by someone in the comments, you want 
$$V = \int_0^8 4-y^{2/3}dy$$
A: Washer method is quite easy. Express the washer area including its hole and extrude it along $y$.
$$
V=\pi\int_{0}^{8}\left(2^2-x^2\right)\,dy=\pi\int_{0}^{2}\left(2^2-x^2\right)\,3x^2 dx= $$
$$
V=3\pi\int_{0}^{2}\left(4-x^2\right)\, \cdot x^2 \, dx = ... = \frac{64 \pi}{5}$$
