Answer
$ \frac{284\pi }{15}$
Work Step by Step
Given
$$ x=1+y^2\ \ \ \ y= x-3\ \ \ \text{about } y-axis$$
First , we find the intersection points
\begin{aligned}
1+y^2&= y+3\\
y^2-y-2&=0\\
(y+1)(y-2)&=0
\end{aligned}
Then
$$y= -1,\ \ \ y=2 $$
Since
$$y+3\geq 1+y^2,\ \ \ \ -1\leq y\leq 2 $$
Then the outer radius $R= y+3$ and the inner $r=1+y^2$
Hence , by using washer method, we have
\begin{aligned}
V &=\pi\int_a^b[R^2-r^2]dy\\
& =\pi \int_0^2\left[(y+3)^2-\left(1+y^2\right)^2\right] dy\\
&= \pi \int_0^2 [-y^4-y^2+6y+8]dy\\
&= \pi \left(-\frac{1}{5}y^5-\frac{1}{3}y^2+3y^2+8y\right)\bigg|_0^2\\
&= \frac{284\pi }{15}
\end{aligned}
