Question

The graph of y = sqrt (16-x^2) ; -2 less than or equal to x less than or equal to 2 is revolved about the x-axis. Find the area of the surface of revolution.

Answer 1

Standard form, \(SA = \int 2\pi y\;ds\) where \(ds = \sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}}\;dx\)

\(\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{16-x^{2}}} = \dfrac{-x}{\sqrt{16-x^{2}}}\)

Can you finish?