Find the area bounded by the curve y = x + 3 ± sqrt(9 +

Find the area bounded by the curve $y = x + 3 \pm \sqrt{9 + 6x - 3x^2}$.

We treat the two branches as the upper and lower edge of one region, then complete the square to reduce the integral to a circular area.

Step 1: Identify the two branches

\[y_+ = x+3+\sqrt{9+6x-3x^2}, \qquad y_- = x+3-\sqrt{9+6x-3x^2}\]

Step 2: Determine the domain

The radical requires $9+6x-3x^2\ge0$, i.e. $x^2-2x-3\le0$, i.e. $(x-3)(x+1)\le0$, so $-1\le x\le3$. At $x=-1$ and $x=3$ the radical vanishes and the two branches meet.

Step 3: Set up the area between the branches

\[A = \int_{-1}^{3}(y_+-y_-)\,dx = \int_{-1}^{3} 2\sqrt{9+6x-3x^2}\,dx\]

Step 4: Complete the square

\[9+6x-3x^2 = -3(x^2-2x-3) = -3\left[(x-1)^2-4\right] = 3\left[4-(x-1)^2\right]\]

So $\sqrt{9+6x-3x^2}=\sqrt3\sqrt{4-(x-1)^2}$.

Step 5: Substitute and recognize a semicircle

Let $u=x-1$: when $x=-1$, $u=-2$; when $x=3$, $u=2$.

\[A = 2\sqrt3\int_{-2}^{2}\sqrt{4-u^2}\,du\]

The integral $\int_{-2}^{2}\sqrt{4-u^2}\,du$ is exactly the area under the upper half of the circle $u^2+v^2=4$ (radius $2$), i.e. the area of a semicircle of radius $2$:

\[\int_{-2}^{2}\sqrt{4-u^2}\,du = \frac{\pi\cdot2^2}{2} = 2\pi\]

Step 6: Calculate the total area

\[A = 2\sqrt3\cdot2\pi = 4\sqrt3\,\pi\]

\[\boxed{A = 4\sqrt3\,\pi \text{ square units}}\]

Step 7: Numerical value

\[A = 4\sqrt3\,\pi \approx 4\times1.732\times3.1416 \approx 21.77 \text{ square units}\]

Summary

\[\boxed{A = 4\sqrt3\,\pi \approx 21.77 \text{ square units}}\]

Key insight: The $\pm$ notation describes an ellipse in disguise: $y_\pm = x+3\pm\sqrt{9+6x-3x^2}$ traces out an ellipse whose axes are tilted relative to the coordinate axes, and completing the square converts the enclosed area directly into $\sqrt3$ times the area of a circle of radius $2$.