Question:
The area of the region bounded by $y^2 = 16 - x^2, y = 0, x = 0$ in the first quadrant is (in square units):
The area of the region bounded by $y^2 = 16 - x^2, y = 0, x = 0$ in the first quadrant is (in square units):
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While you can solve this using the integral $\int_{0}^{4} \sqrt{16-x^2} \, dx$, recognizing the geometric shape of the circle saves significant calculation time and reduces the risk of trigonometric substitution errors.
Updated On: May 2, 2026
- $8\pi$
- $6\pi$
- $2\pi$
- $4\pi$
- $\frac{\pi}{2}$
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The Correct Option is D
Solution and Explanation
Concept:
The equation $y^2 = 16 - x^2$ can be rewritten as $x^2 + y^2 = 16$, which is the standard equation of a circle centered at the origin with radius $r$.
• $r^2 = 16 \Rightarrow r = 4$.
Step 1: Identify the region.
The bounds $y=0$ (x-axis) and $x=0$ (y-axis) in the first quadrant mean we are looking for the area of exactly one-quarter of the circle.
Step 2: Calculate the total area of the circle.
\[ \text{Area}_{\text{circle}} = \pi r^2 = \pi (4)^2 = 16\pi \]
Step 3: Find the area in the first quadrant.
\[ \text{Area}_{\text{quadrant}} = \frac{1}{4} \times 16\pi = 4\pi \]
• $r^2 = 16 \Rightarrow r = 4$.
Step 1: Identify the region.
The bounds $y=0$ (x-axis) and $x=0$ (y-axis) in the first quadrant mean we are looking for the area of exactly one-quarter of the circle.
Step 2: Calculate the total area of the circle.
\[ \text{Area}_{\text{circle}} = \pi r^2 = \pi (4)^2 = 16\pi \]
Step 3: Find the area in the first quadrant.
\[ \text{Area}_{\text{quadrant}} = \frac{1}{4} \times 16\pi = 4\pi \]
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