Q.The surface area of the solid of revolution of the region bounded by $x^2 + y^2 = 4$, $x = -2$ and $x = 2$ about $x$-axis is :

Surface Area of a Solid of Revolution

When a plane curve is rotated about a straight axis, it sweeps out a curved surface, and integral calculus lets us find that surface's area. If a smooth curve $y = f(x)$ from $x = a$ to $x = b$ is revolved about the $x$-axis, the surface area is

The factor $2\pi y$ is the circumference of the circular band traced at height $y$, and $\sqrt{1 + (dy/dx)^2}\,dx$ is the arc-length element $ds$ of the generating curve. Revolving about the $y$-axis instead replaces the radius $y$ with $x$.

For a worked idea, take the line $y = x$ from $x = 0$ to $x = 1$ rotated about the $x$-axis (a cone). Here $dy/dx = 1$, so $\sqrt{1+1} = \sqrt2$ and

$S = \int_0^1 2\pi x \sqrt2\, dx = 2\sqrt2\,\pi \cdot \tfrac{1}{2} = \sqrt2\,\pi,$ …