Calculus

Application of Integrals

Hard

The area bounded by $x = \sqrt{9 - y^2}$ , $x - y + 3 = 0$ and x-axis is :

$\frac{9}{4}(\pi - 2)$ sq. units

$\frac{9}{2}(\pi + 1)$ sq. units

$\frac{9}{4}(\pi + 2)$ sq. units

$\frac{9}{2}(\pi + 2)$ sq. units

✅ Correct Option: 3

$x = \sqrt{9-y^2}$ is the right semicircle of radius 3. Line $y = x+3$ meets x-axis at $(-3,0)$ and y-axis at $(0,3)$ .

Region = triangle (vertices $(-3,0)$ , $(0,0)$ , $(0,3)$ ) plus quarter circle (radius 3, in first quadrant).

Area $= \frac{1}{2}(3)(3) + \frac{1}{4}\pi(3)^2 = \frac{9}{2} + \frac{9\pi}{4} = \frac{9}{4}(\pi + 2)$ .

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