Answer
$$ 4 \pi$$
Work Step by Step
We integrate in order to find the surface area:
$$S= \int_{c}^{d} (2\pi y) \sqrt {1+(y')^2}$$ $$\\ = \int_{1}^{2} (2 \pi ) \sqrt {4y-y^2} \times \sqrt {\dfrac{4}{4(y-1)}}dy \\=4 \pi \int_1^2 (1) dx \\= 4 \pi $$
