<s>[INST] Volume of solid rotated about the x-axis I am to find the volume of the area $R$ bounded by the curve $x=y^2+2$, $y=x-4$ and $y=0$.
I have already found the points of intersection by first setting the lines equal to each other and used the quadratic formula:
\begin{align*}
    y^2+2=y+4
    \\-y^2+y+2=0
    \\
    \\y_{1,2}=\frac{1\pm3}{2}\\
    \\y_1 = 2 \\y_2=-1
\end{align*}
With regard to y:
\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\left[(2\cdot2)+\frac{2^2}{2}-\frac{2^3}{3}\right]-\left[(2\cdot0)-\frac{0^2}{2}-\frac{0^3}{3}\right]\\
A=4+2-\frac{8}{3}\\
A=\frac{10}{3}
\end{align*}
Then I tried finding the volume of $R$:

We have:
$f(y)=2+y-y^2 dy $
\begin{align*}
V= 2\pi\int_a^byf(y) dy\\
=2\pi\left(\int_2^4 y^3+2ydy\right)+2\pi\left(\int_4^6(y^3+2y)-(y^2+4y)dy\right)\\
=4\pi\left(\left[\frac{y^4}{4}+y^2\right]_2^4 + \left[\left(\frac{y^4}{4}+y^2\right)-\left(\frac{y^3}{3}+2y^2\right)\right]_4^6\right)\\
=4\pi \cdot 290\\
= \frac{\pi}{2}(145)
\end{align*}
But the right answer should be $16\frac{\pi}{3}$. What am I doing wrong?
 [/INST] Assuming $R$ is the region in the first quadrant (i.e. with $x\ge0$ and $y\ge0$), then the volume - using cylindrical shells - is
$$\begin{align}
V&=2\pi\int_0^2y((y+4)-(y^2+2))\,\mathrm dy\\[1ex]
&=2\pi\int_0^2(2y+y^2-y^3)\,\mathrm dy\\[1ex]
&=2\pi\left[y^2+\frac{y^3}3-\frac{y^4}4\right]_0^2\\[1ex]
&=2\pi\left(2^2+\frac{2^3}3-\frac{2^4}4\right)=\boxed{\frac{16\pi}3}
\end{align}$$
Your error is in the limits of integration. You seem to be using the bounds for the variable $x\in[2,6]$, not $y\in[0,2]$.
If you did want to set up an integral using the bounds for $x$, you can use disks and washers:
$$\begin{align}
V&=\pi\left(\int_2^4\left(\sqrt{x-2}\right)^2\,\mathrm dx+\int_4^6\left(\left(\sqrt{x-2}\right)^2-(x-4)^2\right)\,\mathrm dx\right)
\end{align}$$
and you would find this to have the same value.
</s>