Question

The area of the region
\(\left\{(x,y) : y² ≤ 8x, y ≥ \sqrt2x, x ≥ 1 \right\}\)
is

• A. \(\frac{13\sqrt2}{6}\)
• B. \(\frac{11\sqrt2}{6}\)
• C. \(\frac{5\sqrt2}{6}\)
• D. \(\frac{19\sqrt2}{6}\)

Answer 1

The Correct Option is B

To find the area of the region defined by the inequalities \(\left\{(x,y) : y^2 \leq 8x, y \geq \sqrt{2}x, x \geq 1 \right\}\), we need to understand the geometric meaning of these inequalities and compute the area accordingly.

1. The inequality \(y^2 \leq 8x\) represents the region inside or on the parabola \(y = \sqrt{8x}\) and \(y = -\sqrt{8x}\). However, since \(y \geq \sqrt{2}x\), the relevant part is only above the line \(y = \sqrt{2}x\).

2. The inequality \(y \geq \sqrt{2}x\) restricts the region to be above the line passing through the origin with slope \(\sqrt{2}\).

3. Additionally, \(x \geq 1\) restricts the region to the right of \(x = 1\).

4. We find the points of intersection for \(y = \sqrt{8x}\) and \(y = \sqrt{2}x\).

   Equating, \(\sqrt{8x} = \sqrt{2}x \Rightarrow 8x = 2x^2\).

   This simplifies to \(2x^2 - 8x = 0 \Rightarrow 2x(x - 4) = 0\).

   So, the points are \(x = 0\) and \(x = 4\).

5. The area to be computed is from \(x = 1\) to \(x = 4\).

   The integral of the upper curve − lower line from 1 to 4 gives the desired area.

   Area = \(\int_{1}^{4} (\sqrt{8x} - \sqrt{2}x)\, dx\)

6. Calculate the integral:

   \(\int \sqrt{8x} \, dx = \int \sqrt{8}\sqrt{x} \, dx = \frac{2\sqrt{8}}{3}x^{3/2}\)

   \(\int \sqrt{2}x \, dx = \frac{\sqrt{2}}{2}x^2\)

   Perform the computation:

   \(\left[ \frac{2\sqrt{8}}{3}x^{3/2} \right]_{1}^{4} - \left[ \frac{\sqrt{2}}{2}x^2 \right]_{1}^{4}\)

7. Substitute and evaluate the limits:

   \(\frac{16}{3}\sqrt{2} - \sqrt{2}\) = \(\frac{11 \sqrt{2}}{3}\). Simplifying further gives:

   Final Area = \(\frac{11\sqrt{2}}{6}\).

The answer is the choice \(\frac{11\sqrt{2}}{6}\).