## Thomas' Calculus 13th Edition

$4 \pi a^2$
Our aim is to integrate the integral to compute the surface area. In order to solve the integral, we have: $Surface \space Area(S_A)= (2 \pi)\int_{a}^{b} y \sqrt {1+(\dfrac{dy}{dx})^2}$ or, $=(2 \pi)\int_{-a}^{a} \sqrt {a^2-x^2} \times \sqrt {\dfrac{a^2}{a^2-x^2} } dx$ or, $=(2 \pi)\int_{-a}^{a} a dx$ or, $=2 \pi a (x)_{-a}^a$ or, $=4 \pi a^2$