Answer
$$\pi$$
Work Step by Step
Let us consider that $f(x)$ is a non-negative function which defines a continuous first derivative on the interval $[a,b]$. The area of the surface obtained when the graph of a function $f(x)$ on the interval $\left[ {a,b} \right]$ is revolved about the $x$- axis can be expressed as:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
We have: $f(y)=x=\sqrt {2y-y^2} \implies f'(y)=\dfrac{1-y}{\sqrt {2y-y^2}}$
Use formula: $\int x^n dx=\dfrac{x^{n+1}}{n+1}$
Then $$S=\int_{1}^{3/2} 2\pi (\sqrt {2y-y^2}) \sqrt {1 + (\dfrac{1-y}{\sqrt {2y-y^2}})^2} dy \\=2 \pi \int_{1}^{3/2} [2y-y^2+(1-y)^2]^{1/2} \ dy \\=2 \pi \int_{1} ^{3/2} (1) \ dy \\=2\pi [y] _1^{3/2} \\ =\pi$$
