My question is how do I calculate the volume of the region between $f(x) = (2x+1)\sqrt{x^2 + x}\,$, $\,g(x) = x^2\,$, and $x = 1$ using the Shell Method?
Here are the 3D visuals I graphed using Geogebra.
I've already solved it using the Disc/Washer Method:
$$ I_\text{Disc/Washer} = \int_{x=0}^{x=1} \pi \left[\left((2x + 1)\sqrt{x^2 + x}\right)^2 - \left(x^2\right)^2\right] \,\mathrm{d}x = \frac{143\pi}{30} $$
I think the answer is already correct but I still would want to be able to solve it using the Shell Method, too, if possible. Here's what I've tried so far by using an inverse branch of $\,(2x+1)\sqrt{x^2+x}\,$.
$$ I_\text{Shell} = \int_{y=0}^{y=3\sqrt{2}} 2\pi y\left[\sqrt{y} - \left(-\frac{1}{2}+\frac{\sqrt{\sqrt{16x^{2}+1}+1}}{2\sqrt{2}}\right)\right] \,\mathrm{d}y $$
The result is $\,\dfrac{\pi}{30}\left(432\sqrt[4]{2}\sqrt{3}-391\right)\,$ according to WolframAlpha which doesn't seem to match up with $\,I_\text{Disc/Washer}\,$ but I don't know where I made a mistake. Did I get the lower and upper limits wrong? Thank you.
You need to split the last integral into two regions. The integrand has your form up to $y=1$. Above that value, the length of the shell is bounded by $x=1$, not by $x=\sqrt y$.
$$ I_\text{Shell} = \int_{y=0}^{y=1} 2\pi y\left[\sqrt{y} - \left(-\frac{1}{2}+\frac{\sqrt{\sqrt{16x^{2}+1}+1}}{2\sqrt{2}}\right)\right] \,\mathrm{d}y +\int_{y=1}^{y=3\sqrt{2}} 2\pi y\left[1 - \left(-\frac{1}{2}+\frac{\sqrt{\sqrt{16x^{2}+1}+1}}{2\sqrt{2}}\right)\right] \,\mathrm{d}y$$