Finding volume of solid obtained by rotating the regin enclosed by $y=x^3, y=0, x=1$ about $x=2$ line is
What i try:: 
Volume of solid
$$V=\pi\int^{8}_{0}\bigg[(2-x)^2-1\bigg]dy$$
$$V=\pi\int^{8}_{0}\bigg[(2-y^{\frac{1}{3}})^2-1\bigg]dy=-\frac{24\pi}{5}$$
Whats wrong with my solution, please Help me. Thanks
A simple method to calculate the volume is to use the Pappus theorem. The area of the region of interest is $\int_0^1 x^3 dx=1/4$. The $x$-coordinate $X_c$ of the centroid of a region bounded by two curves $f(x)$ and $g(x)$ in the interval $[a,b]$ is given by the formula
$$\displaystyle X_c= {\frac{{\large\int_a^b {x\Big[ {f\left( x \right) – g\left( x \right)} \Big]dx} }}{{\large\int_a^b {\Big[ {f\left( x \right) – g\left( x \right)} \Big]dx} }}}$$
In our case, $f(x)=x^3$, $g(x)=0$, and $[a,b]=[0,1]$. Making these substitutions in the formula, the integral in the numerator gives $1/5$ and that in the denominator gives $1/4$, so that $X_c=4/5$.
Now we can apply the Pappus formula to get
$$V=\frac 14 \, 2\, \pi \, \frac{6}{5}=\frac {3\pi}{5}$$
The same result is obtained using alternatively your method of integration. The correct bounds are $[0,1]$, so that
$$V=\pi\int^{1}_{0}\bigg[(2-y^{\frac{1}{3}})^2-1\bigg]dy=\frac{3\pi}{5}$$