Find the volume of region bounded by $y=x^2 +4$, $y=2x+3$, and $x=0$, that revolves about the $y$ axis.
I choose to integrate on the $x$ axis from $0$ to $1$ and from heropup's answer (Solids of Revolution Question (Method of Cylinders vs Disc/Washers)), I choose the method of cylindrical shell as the axis of rotation is perpendicular to the interval of integration.
$$\int_0^1 2\pi x (x^2 + 4 - 2x-3) dx = \frac{\pi}{6}.$$
I was wondering, whether could I integrate on the $y$ axis instead, and from the region bounded, I need to integrate from $3$ to $5$ and since now the axis of rotation is parallel to the interval of integration, we need to use the method of rings (cross sectional area). Will this yield the same volume through integration and is my thoughts correct? Or am I only restricted to using the method of cylindrical shells for this question?
Have an exam tmr and I am worried between these 2 methods and whether I need to pick the correct one or whether it is flexible between the 2 depending on axis of rotation and interval of integration (x or y axis).
Yes, it can be done.
If $3\leqslant y\leqslant4$, then you are only interest in those points $(x,y)$ with $0\leqslant x\leqslant\frac{y-3}2$. And if $4\leqslant y\leqslant5$, then you are interested in those points $(x,y)$ with $\sqrt{y-4}\leqslant x\leqslant\frac{y-3}2$. So, your volume is\begin{align}\pi\left(\int_3^4\left(\frac{y-3}2\right)^2\,\mathrm dy+\int_4^5\left(\frac{y-3}2\right)^2-(y-4)\,\mathrm dy\right)&=\pi\left(\frac1{12}+\frac1{12}\right)\\&=\frac\pi6.\end{align}