The area (in sq. units) of the smaller region bounded by the circle x…

The correct answer is Option (3) → \(\displaystyle\frac{9}{4}(\pi-2)\)

The line x + y = 3 meets the circle (radius 3) at (3, 0) and (0, 3). The smaller region lies in the first quadrant, between the circle above and the line below, for \(0 \le x \le 3\):

    \( \displaystyle A = \int_{0}^{3}\sqrt{9-x^{2}}\,dx - \int_{0}^{3}(3-x)\,dx \)

The first integral is the quarter of the circle in the first quadrant:

    \( \displaystyle \int_{0}^{3}\sqrt{9-x^{2}}\,dx = \left[\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3} \)

    \( \displaystyle = \frac{9}{2}\times\frac{\pi}{2} = \frac{9\pi}{4} \)

The second integral is the triangle under the line:

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

So

    \( \displaystyle A = \frac{9\pi}{4}-\frac{9}{2} = \frac{9}{4}(\pi-2) \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{9\pi}{4}\) — Incorrect. This is the whole quarter circle; the triangle under the line is not taken away.

• Option (2) \(\displaystyle\frac{9}{2}(\pi-2)\) — Incorrect. This is twice the correct value, \(\frac{9\pi}{2}-9\).

• Option (4) \(\displaystyle\frac{9}{4}(\pi-1)\) — Incorrect. This takes away \(\frac{9}{4}\) instead of the triangle’s area \(\frac{9}{2}\).