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Prove It Since you are rotating around the $\displaystyle \displaystyle y$ axis, this will be a $\displaystyle \displaystyle dy$ integral.
The graphs intersect where the equations are equal...
$\displaystyle \displaystyle x = 1 + (x-3)^2$
$\displaystyle \displaystyle x = 1 + x^2 - 6x + 9$
$\displaystyle \displaystyle 0 = x^2 - 7x + 10$
$\displaystyle \displaystyle 0 = (x - 2)(x - 5)$
$\displaystyle \displaystyle x = 2$ or $\displaystyle \displaystyle x = 5$.
When $\displaystyle \displaystyle x = 5, y = 2$, so they intersect at $\displaystyle \displaystyle (x,y) = (5, 2)$.
The volume is calculated by rotating the area around the $\displaystyle \displaystyle y$ axis.
This area needs to be thought of as a series of rectangles, which when rotated, become cylinders.
The length of each rectangle is $\displaystyle \displaystyle (y + 3) - (1 + y^2) = 2 + y - y^2$ and the width is $\displaystyle \displaystyle dy$, a small change in $\displaystyle \displaystyle y$.
When rotated, each cylinder has a radius the same as the length of each rectangle, so the circular cross-sectional area is $\displaystyle \displaystyle \pi r^2 = \pi (2 + y - y^2)^2$. The height of each cylinder is the same as the width of each rectangle, $\displaystyle \displaystyle dy$.
Therefore the volume of each cylinder is $\displaystyle \displaystyle \pi (2 + y - y^2)^2\,dy$.
So the entire volume can be approximated by $\displaystyle \displaystyle \sum{\pi(2+y-y^2)^2\,dy}$, and when you increase the number of cylinders, this sum converges on an integral and the approximation becomes exact.
So $\displaystyle \displaystyle V = \int_0^2{\pi (2 + y - y^2)^2\,dy}$.