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?

I don't see why you need to split the region. Notice that $y = 4x^2$ and $y = 6 - 2x$ both intercept at: $$4x^2 = 6-2x \implies2x^2 + x - 3 =0 \implies x = \frac{-1\pm\sqrt{1+24}}{4} \implies x = \frac{-1\pm 5}{4},$$ so $x = -3/2$ and $x = 1$. Since the area of revolution of $y = f(x)$ around the $x$ axis is $$A = \pi \int_{a}^b f(x)^2 \ {\rm d}x$$ and $6 - 2x > 4x^2$ in $(-3/2, 1)$, the area you're looking for is: $$A = \pi\int_{-3/2}^1 (6-2x)^2 \ {\rm d}x - \pi \int_{-3/2}^1 (4x^2)^2 \ {\rm d}x.$$
