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The area of the region bounded by the curve y=9−x2y=\sqrt{9-x^2} and x-axis is

Solution

✅ Correct Option: 4

y=9−x2y=\sqrt{9-x^2} is the upper half of the circle x2+y2=9x^2+y^2=9 of radius 3. With the x-axis it bounds a semicircle, whose area is 12π(3)2=9π2\frac12\pi(3)^2=\frac{9\pi}{2}.

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