Question

The area of the region \[ A = \{(x,y) : 4x^2 + y^2 \le 8 \;\text{and}\; y^2 \le 4x\} \] is

• A. \( \dfrac{\pi}{2} + 2 \)
• B. \( \pi + 4 \)
• C. \( \pi + \dfrac{2}{3} \)
• D. \( \dfrac{\pi}{2} + \dfrac{1}{3} \)

Hint:

When a region is bounded by two curves, always identify intersection points first, then integrate with respect to the variable that simplifies the limits.

Answer 1

The Correct Option is C

To determine the area of the region given by:

\(A = \{(x, y) : 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\}\)

we need to analyze the inequalities separately.

The first inequality is:

1. \(4x^2 + y^2 \le 8\)

This can be rewritten in standard ellipse form:

\(\frac{x^2}{2} + \frac{y^2}{8} \le 1\)

Thus, it represents an ellipse centered at the origin (0, 0) with semi-major axis of length 2 along the \(x\)-axis and semi-minor axis of length \(\sqrt{8} = 2\sqrt{2}\) along the \(y\)-axis.

The second inequality is:

1. \(y^2 \le 4x\)

This represents a parabola that opens to the right.

We need to find the area where these two regions intersect.

To find the points of intersection, set \(4x^2 + y^2 = 8\) and \(y^2 = 4x\) equal and solve for \(x\) and \(y\):

Substitute \(y^2 = 4x\) into the ellipse equation:

1. \(4x^2 + 4x = 8\)

Simplifying gives:

1. \(x^2 + x - 2 = 0\)

Factoring gives:

1. \((x - 1)(x + 2) = 0\)

So, \(x = 1\) or \(x = -2\).

For \(x=1\), \(y^2 = 4 \times 1 = 4\) gives \(y = 2\) or \(y = -2\).

For \(x=-2\), \(y^2 = 4 \times (-2) = -8\) is not valid, thus we only use \(x = 1\).

The area enclosed by the parabola from \(x=0\) to \(x=1\) can be calculated using integration:

Area under \(y^2 = 4x\):

1. \(\int_0^1 \sqrt{4x} \, dx = \int_0^1 2\sqrt{x} \, dx\)

After integrating:

1. \(= \left. \frac{4}{3}x^{3/2} \right|_0^1 = \frac{4}{3}\)

Area of the quarter ellipse \(4x^2 + y^2 = 8\):

1. \(\frac{1}{4} \cdot \pi \cdot 2 \cdot 2\sqrt{2} = \frac{\pi}{2}\)

The total intersection area:

1. \(\text{Total Area} = \frac{\pi}{2} + \frac{4}{3}\)

Thus, the area of the region is:

1. \(\pi + \frac{2}{3}\)

Therefore, the correct answer is:

\( \pi + \frac{2}{3} \)