Volume of solid of revolution by shell method consider the region bounded by $ \displaystyle y=4{{x}^{2}}$ and $ \displaystyle 2x+y=6$. What is the volume of solid of revolution about $\displaystyle x$-axis.
What is thought about setting the integral:
I split the region into two parts
$\displaystyle V=4\pi \int\limits_{0}^{4}{y\left( 1-\frac{\sqrt{y}}{2} \right)\,dy}+2\pi \int\limits_{4}^{9}{y\left( \frac{6-y}{2}-\frac{\sqrt{y}}{2} \right)\,dy}$
Will it work?

Below is the plot for $y = 4x^2$, and $y=6-2x:$

There is absolutely no need to separate $\mathfrak{R}$ into two regions. Find the intersection of the two curves and use $r = r_{out} - r_{in}$. Recall that 
$$
\mathfrak{R} = \int_{x_0}^{x_1} \pi r^2 \ dr.
$$
where $x_0$ is the leftmost intersection and $x_1$ is the rightmost.
