The area bounded by x=√(9-y^2) , x-y+3=0 and x-axis is:

The correct answer is Option (3) → \(\displaystyle\frac{9}{4}(\pi+2)\) sq. units

The curve \(x = \sqrt{9-y^2}\) is the right half (x ≥ 0) of the circle \(x^2 + y^2 = 9\).

The line x − y + 3 = 0, that is x = y − 3, meets the x-axis at (−3, 0) and the circle at (0, 3).

So the region lies above the x-axis, from y = 0 to y = 3, with the line on the left and the arc on the right. Using horizontal strips:

    \( \displaystyle \text{Area} = \int_{0}^{3}\left[\sqrt{9-y^2} - (y-3)\right]dy \)

The first integral is the area of a quarter of the circle of radius 3:

    \( \displaystyle \int_{0}^{3}\sqrt{9-y^2}\,dy = \frac{1}{4}\pi(3)^2 = \frac{9\pi}{4} \)

The second integral:

    \( \displaystyle \int_{0}^{3}(3-y)\,dy = \left[3y - \frac{y^2}{2}\right]_{0}^{3} = \frac{9}{2} \)

Hence

    \( \displaystyle \text{Area} = \frac{9\pi}{4} + \frac{9}{2} \)

    \( \displaystyle = \frac{9}{4}(\pi+2) \text{ sq. units} \)

This is the quarter circle in the first quadrant together with the triangle with vertices (0, 0), (−3, 0) and (0, 3).