Q118 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
The area of the region (in sq units) bounded by the curve $
y=\sqrt{16-x^2} \text { and } x \text {-axis is }$
- A256$\pi$
- B16$\pi$
- C20$\pi$
- Correct answerD8$\pi$
Solution
The given equation is $y = \sqrt{16 - x^2}$. Squaring both sides, we get $y^2 = 16 - x^2$, which implies $x^2 + y^2 = 16$.
This is the equation of a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{16} = 4$.
Since $y = \sqrt{16 - x^2}$, $y$ must be non-negative, meaning the curve represents the upper semi-circle.
The area of a full circle is $\pi r^2$. For a circle with radius $r = 4$, the area is $\pi (4)^2 = 16\pi$.
The area of the upper semi-circle bounded by the $x$-axis is half the area of the full circle, which is $\dfrac{1}{2} \times 16\pi = 8\pi$.
Answer: 8$\pi$
This is the equation of a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{16} = 4$.
Since $y = \sqrt{16 - x^2}$, $y$ must be non-negative, meaning the curve represents the upper semi-circle.
The area of a full circle is $\pi r^2$. For a circle with radius $r = 4$, the area is $\pi (4)^2 = 16\pi$.
The area of the upper semi-circle bounded by the $x$-axis is half the area of the full circle, which is $\dfrac{1}{2} \times 16\pi = 8\pi$.
Answer: 8$\pi$
Hence, the correct answer is option: 'D'.