How do I use the shell method to calculate the volume of this region between two solids of revolution?
Solution 1:
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 $x=\sqrt y$.
$$ I_\text{Shell} = \int_{y=0}^{y=1} 2\pi y\left[\sqrt{y} - \left(-\frac{1}{2}+\frac{\sqrt{\sqrt{16x^{2}+1}+1}}{2\sqrt{2}}\right)\right] \,\mathrm{d}y +\int_{y=1}^{y=3\sqrt{2}} 2\pi y\left[1 - \left(-\frac{1}{2}+\frac{\sqrt{\sqrt{16x^{2}+1}+1}}{2\sqrt{2}}\right)\right] \,\mathrm{d}y$$