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 $277π/3$.

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:
$$\begin{align}
V=&2\pi\int_1^5\sqrt{y-1}y\,dy+2\pi\int_2^5(3-(1+\sqrt{6-y}))y\,dy
\\&+2\pi\int_5^{10}(3-\sqrt{y-1})y\,dy+
2\pi\int_5^6((1+\sqrt{6-y})-(1-\sqrt{6-y}))y\,dy=\frac{277\pi}{3}
\end{align}$$

