Question

The area bounded by the curve 
\(y=|x^2−9| \)
and the line y = 3 is

• A. \(4(2\sqrt3+\sqrt6-4)\)
• B. \(4(4\sqrt3+\sqrt6-4)\)
• C. \(8(4\sqrt3+3\sqrt6-9)\)
• D. \(8(4\sqrt3+\sqrt6-9)\)

Answer 1

The Correct Option is A

 To find the area bounded by the curve \(y = |x^2 - 9|\) and the line \(y = 3\), we follow these steps:

Understand the function \(y = |x^2 - 9|\). This can be broken down into two cases:

• For \(x^2 \leq 9\), \(y = 9 - x^2\).
• For \(x^2 > 9\), \(y = x^2 - 9\).

Identify the points of intersection between the curves \(y = |x^2 - 9|\) and the line \(y = 3\). Setting the equations equal gives us:

• \(|x^2 - 9| = 3 \implies x^2 - 9 = 3\) or \(-(x^2 - 9) = 3\).
• Solving \(x^2 - 9 = 3\) gives \(x^2 = 12 \Rightarrow x = \pm 2\sqrt{3}\).
• Solving \(-(x^2 - 9) = 3\) gives \(x^2 = 6 \Rightarrow x = \pm \sqrt{6}\).

Calculate the area between the curves. We do this by integrating the area under the curves between these points:

• From \(x = -2\sqrt{3}\) to \(x = -\sqrt{6}\), the curve is \(y = 9 - x^2\).
• From \(x = -\sqrt{6}\) to \(x = \sqrt{6}\), the curve is \(y = x^2 - 9\).
• From \(x = \sqrt{6}\) to \(x = 2\sqrt{3}\), the curve is again \(y = 9 - x^2\).

Perform the integration over the respective intervals:

• \(\int_{-2\sqrt{3}}^{-\sqrt{6}} ((9 - x^2) - 3) \, dx = \int_{-2\sqrt{3}}^{-\sqrt{6}} (6 - x^2) \, dx\)
• \(\int_{-\sqrt{6}}^{\sqrt{6}} ((x^2 - 9) - 3) \, dx = \int_{-\sqrt{6}}^{\sqrt{6}} (x^2 - 12) \, dx\)
• \(\int_{\sqrt{6}}^{2\sqrt{3}} ((9 - x^2) - 3) \, dx = \int_{\sqrt{6}}^{2\sqrt{3}} (6 - x^2) \, dx\)
• \(A_1 = \int_{-2\sqrt{3}}^{-\sqrt{6}} (6 - x^2) \, dx + \int_{\sqrt{6}}^{2\sqrt{3}} (6 - x^2) \, dx\)
• \(A_2 = \int_{-\sqrt{6}}^{\sqrt{6}} (x^2 - 12) \, dx\)

Combine the results to find the total area:

The total area \(A = A_1 + A_2 = 4(2\sqrt{3} + \sqrt{6} - 4)\), which matches the given correct answer.