Volume of $y=x^2+1; y=-x^2+2x+5; x=0; x=3$ about $x$ axis (Shell Method). I was working on this exercise for an assignment. However, I get stuck in the following part.

$
y=-x^2+2x+5
$

Complete the square

$y=-(x^2-2x)+5$

$(b/2)^2=(-2/2)^2=1$

$y=-(x^2-2x+1-1)+5$

$y=-(x^2-2x+1)+5+1$

$y=-(x-1)^2+6$

$y-6=-(x-1)^2$

$\sqrt{6-y}+1=x$

For $x=0$, $\sqrt{6-y}+1=x$ has no real answer and I was wondering, is it possible to solve this with the Shell Method about the $x$-axis? I tried with Washer Method, and the expected volume is $277π/3$.
 A: Yes, it can be solved through the shell method. Consider the picture below:

Let us first see what is the volume of the region for which $0\leqslant x\leqslant2$. The shell method tells us that it is equal to\begin{multline}2\pi\left(\int_1^5y\sqrt{y-1}\,\mathrm dy+\int_5^6y\left(\left(1+\sqrt{6-y}\right)-\left(1-\sqrt{6-y}\right)\right)\,\mathrm dy\right)\\=2\pi\left(\int_1^5y\sqrt{y-1}\,\mathrm dy+\int_5^62\sqrt{6-y}\,\mathrm dy\right)=\frac{152}3\pi.\end{multline}And the volume of the region for which $2\leqslant x\leqslant3$ is (again, by the shell method)$$2\pi\left(\int_2^5y\left(3-\left(1+\sqrt{6-y}\right)\right)\,\mathrm dy+\int_5^{10}y\left(3-\sqrt{y-1}\right)\,\mathrm dy\right)=\frac{125}3\pi.$$And $\dfrac{152}3\pi+\dfrac{125}3\pi=\dfrac{177}3\pi$.
A: Note that in $[0,3]$, $-x^2+2x+5\geq x^2+1$ for $x\in [0,2]$. Therefore, by the Washer Method, the volume is given by
$$V=\pi\int_0^2((-x^2+2x+5)^2-(x^2+1)^2)\,dx+\pi\int_2^3((x^2+1)^2-(-x^2+2x+5)^2)\,dx=\frac{277\pi}{3}$$
Using the Shell Method is a bit more complicated because we have to split the evaluation into 4 pieces:
$$\begin{align}
V=&2\pi\int_1^5\sqrt{y-1}y\,dy+2\pi\int_2^5(3-(1+\sqrt{6-y}))y\,dy
\\&+2\pi\int_5^{10}(3-\sqrt{y-1})y\,dy+
2\pi\int_5^6((1+\sqrt{6-y})-(1-\sqrt{6-y}))y\,dy=\frac{277\pi}{3}
\end{align}$$

