Question

What is the surface area of the solid obtained by revolving the curve \(y=\sqrt{9-x^2}\), \(-2\le x\le 2\), about the x-axis?

• A. \(18\pi\)
• B. \(22\pi\)
• C. \(24\pi\)
• D. \(30\pi\)

Hint:

The arc lies on a sphere of radius 3; use \(S=2\pi Rh\) for a spherical zone.

Answer 1

The Correct Option is C

Step 1: The curve \(y=\sqrt{9-x^2}\) is the upper half of the circle \(x^2+y^2=9\), a circle of radius \(R=3\) centered at the origin. Revolving an arc of a circle about a diameter (the x-axis here) produces a spherical zone, a band cut from the surface of a sphere of radius \(R\).

Step 2: By Archimedes' theorem on spherical zones, the lateral surface area of a zone depends only on the radius of the sphere and the height of the zone along the axis, not on its position: \[ S = 2\pi R h \] where \(h\) is the length of the interval on the axis of revolution.

Step 3: Here \(R=3\) and the zone runs from \(x=-2\) to \(x=2\), so \[ h = 2-(-2) = 4 \] \[ S = 2\pi(3)(4) = 24\pi \] \[ \boxed{S = 24\pi} \]