Finding the volume of a solid through 2$\pi$ radians With reference to this question
Curve $C$ with the equation:
$$y = (2x-1)^{\frac{3}{4}}, \quad x \ge\frac{1}{2}$$
The finite region $S$, is bounded by the curve $C$, the $x$-axis. The $y$-axis and the line $y = 8$. This region is rotated through $2\pi$ radians about the $x$-axis.
Find the exact value of the volume of the solid generated.
Here's what I have tried:
$\mathbf{k}$ or $\mathbf{x}$ was found to be at $\frac{17}{2}$
$$\int\limits^{17/2}_{1/2}(2x-1)^{3/4}dx=\frac{256}{7}$$
I'm wondering if I approached this correctly?

Curve $C$ is given by $y = (2x-1)^{\frac{3}{4}}$ for $x \ge\frac{1}{2}$.
Note that the region S, is bounded by the curve C, the x-axis, the y-axis and the line $y = 8$. This region is rotated through $2\pi$ about the x-axis.
a) If we find the volume using shell method, it is one integral.
$\displaystyle y = (2x-1)^{\frac{3}{4}} \implies x = \frac{y^{4/3} + 1}{2}$
So the integral can be set up as,
$V = \displaystyle \int_0^8 \int_0^{(1 + y^{4/3})/2} 2 \pi \ y \ dx \ dy = \frac{1696 \pi}{5}$
b) If we find volume using washer method, we will have to split the integral into two, which is between $0 \leq x \leq \frac{1}{2}$ and between $\frac{1}{2} \leq x \leq \frac{17}{2}$. The volume for the first part will be a cylinder of radius $8 (0 \leq y \leq 8)$ whereas the second part will have radius as $(2x-1)^{\frac{3}{4}} \leq y \leq 8$.
$V = \displaystyle \int_{1/2}^{17/2} \int_{(2x-1)^{3/4}}^8 2\pi \ y \ dy \ dx + \int_0^{1/2} \int_0^8 2\pi \ y \ dy \ dx$
