Question: the area enclosed by $\displaystyle y=x^2-6x+18 $ and y =10 is rotated about the x-axis; find the volume generated. take an element of area parallel to the x-axis of length$\displaystyle (x_{2}-x_{1}) $ express the typical elements of volume in terms of y by using the fact that $\displaystyle x_{1} $ and $\displaystyle x_{2} $ are the roots of $\displaystyle x^2 -6x + (18-y)= 0 $

my attempt:

the graph is

$\displaystyle y=x^2-6x+18 $

when x is the subject

$\displaystyle x = 3+(y-9)^{0.5} $

the interval is y =9 to y=10

the minimum point is (3,9)

$\displaystyle v= \int_{9}^{10} \pi (3+(y-9)^{0.5})^{2} \mathrm{d}y $

$\displaystyle v = 13.5000002 \pi $

please help me with this i am finding it difficult to understand how to tackle the book gives the answer as $\displaystyle 8\pi $