Question

The area bounded by the parabola \( y^2 = x \), the straight line \( y = 4 \) and \( Y \) axis is

• A. \( 2\sqrt{7} \) sq. units
• B. \( \frac{64}{3} \) sq. units
• C. \( \frac{16}{3} \) sq. units
• D. \( 7\sqrt{2} \) sq. units

Hint:

When the bounding curves are given as \(y = f(x)\) or \(x = g(y)\), choose the variable that makes the integration simpler. Here, integrating with respect to \(y\) avoids splitting the region.

Answer 1

The Correct Option is B

Step 1: Understanding the Question: We need the area of the region bounded by the parabola \(y^2 = x\) (opening to the right), the horizontal line \(y = 4\), and the Y-axis (\(x = 0\)).

Step 2: Key Formula or Approach: Since the region is bounded on the left by \(x=0\) and on the right by the parabola \(x = y^2\), it is convenient to integrate with respect to \(y\). The limits of \(y\) are from the intersection of the parabola with the Y-axis (\(y=0\)) up to the given line \(y=4\).

Step 3: Detailed Explanation: The curve \(y^2 = x\) meets the Y-axis at the origin \((0,0)\). The line \(y=4\) cuts the parabola at \(x = 4^2 = 16\). For a fixed \(y\) between \(0\) and \(4\), the \(x\)-coordinate on the parabola is \(x = y^2\), while the left boundary is \(x=0\). Thus the area is: \[ \text{Area} = \int_{y=0}^{4} \bigl( y^2 - 0 \bigr) \, dy = \int_{0}^{4} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{4} = \frac{64}{3} \text{ square units}. \]

Step 4: Final Answer: Option (B) is correct.