Answer
$\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $
Work Step by Step
Surface area is given by:
$S=\iint_{D} \sqrt {1+\dfrac{a^2x^2}{x^2+y^2}+\dfrac{a^2y^2}{x^2+y^2}}$
or, $=\iint_{D} \sqrt {1+\dfrac{a^2x^2+a^2y^2}{x^2+y^2}}$
or, $=\sqrt {1+a^2} \iint_{D} dA$
Since, $(\dfrac{1}{a})^2 \leq x^2+y^2 \leq (\dfrac{2}{a})^2 $
So, $S= \sqrt {1+a^2} \iint_{D} dA=\sqrt {1+a^2} \times\pi (\dfrac{2}{a})^2 -\pi (\dfrac{1}{a})^2=\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $
