Answer
$\dfrac{8}{15}$
Work Step by Step
Here, we have the area function from the point $y=0$ to $y=1$ is given as: $A_1(y)=(y^2)^2=y^4$ and the area function from the point $y=1$ to $y=2$ is given as: $A_1(y)=(2-y)^2$
Thus, the total volume by the slicing method is:
$A(x)=\int_0^1 y^4 \ dy+ \int_1^2 (2-y)^2 \ dy$
or, $=[\dfrac{y^5}{5}]_0^1-[\dfrac{(2-y)^3}{3}]_1^2$
or, $=\dfrac{(1)^5}{5}+ \dfrac{1}{3}$
or, $=\dfrac{8}{15}$
