Q.The area of the region bounded by the curve $y = \sqrt{16 - x^2}$ and x-axis is
(A) $8$ sq units
(B) $20\pi$ sq units
(C) $16\pi$ sq units
(D) $256\pi$ sq units

Idea: Squaring $y=\sqrt{16-x^2}$ gives $x^2+y^2=16$ with $y\ge 0$ — the upper half of a circle of radius $4$. Bounded below by the $x$-axis, the region is a semicircle. …

Identify the curve

Square both sides: $y^2=16-x^2$, i.e. $x^2+y^2=16$. Because $y=\sqrt{16-x^2}\ge 0$, only the upper half is taken — this is the top semicircle of the circle of radius $4$ centred at the origin, running from $x=-4$ to $x=4$.

Set up the area

The $x$-axis ($y=0$) closes the region below, so we want the semicircular area:

Evaluate

Using $\displaystyle\int\sqrt{16-x^2}\,dx=\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac{x}{4}$:

This is just the area of a semicircle of radius $4$: $\tfrac12\pi(4)^2=8\pi$. …

Method: Area under a square-root curve of the form $y = \sqrt{a^2 - x^2}$

Whenever the curve is $y = \sqrt{a^2 - x^2}$, recognise that squaring gives $x^2 + y^2 = a^2$ with $y \ge 0$ — the upper half of a circle of radius $a$. The area it bounds with the $x$-axis is therefore a semicircle.

Steps

Step 1: Identify the curve.

Square both sides to reveal the circle $x^2 + y^2 = a^2$; the positive square root keeps only the top half. Read off the radius $a$.

Step 2: Set the limits.

The semicircle runs from $x = -a$ to $x = a$, where it meets the $x$-axis.

Step 3: Evaluate the integral (or use geometry). …

Common Mistakes

Mistake 1: Computing the whole circle's area, $16\pi$.

Why it's wrong: $y=\sqrt{16-x^2}$ is only the upper half of $x^2+y^2=16$ (since $y\ge 0$), so together with the $x$-axis it encloses a semicircle, not the full disc. Taking $16\pi$ (the tempting distractor) doubles the true value. Correct approach: use half the circle area, $\tfrac12\pi(4)^2=8\pi$. …