Question:medium
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
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