Question:medium

The area in sq. units bounded by the curve \( y = 2\sqrt{1 - x^2} \) and the x-axis is :

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The graph \( y=\sqrt{a^2-x^2} \) represents the upper semicircle. Similarly, \( y=b\sqrt{1-\frac{x^2}{a^2}} \) represents the upper half of an ellipse. Recognizing standard curves can save lengthy integration.
Updated On: May 29, 2026
  • \( \pi \)
  • \( 2\pi \)
  • \( \pi/2 \)
  • \( 4\pi \)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
Area calculation in coordinate geometry can be performed using definite integrals, but identifying standard geometric shapes can often provide a more direct path.
The equation \( y = 2\sqrt{1 - x^2} \) belongs to a family of curves derived from the standard equations of conic sections, specifically ellipses and circles.
Since the square root symbol represents the principal (non-negative) root, the value of \( y \) is always greater than or equal to zero for all real values of \( x \).
This indicates that the curve represents only the portion of the geometric figure that lies on or above the x-axis.
The boundaries of the region are the curve itself and the x-axis (\( y = 0 \)).
Step 2: Key Formula or Approach:
The standard equation of an ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The total area of a complete ellipse is given by:
\[ \text{Total Area} = \pi a b \]
Where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes.
Since we are dealing with the region bounded by the curve and the x-axis for \( y \geq 0 \), we are looking for the area of exactly half of the ellipse.
Step 3: Detailed Explanation:
Let's analyze the given equation to identify the parameters of the ellipse.
Given: \( y = 2\sqrt{1 - x^2} \)
To remove the radical sign, we square both sides of the equation:
\[ y^2 = (2\sqrt{1 - x^2})^2 \]
\[ y^2 = 4(1 - x^2) \]
Now, distribute the 4 into the terms inside the parentheses:
\[ y^2 = 4 - 4x^2 \]
Move the term involving \( x \) to the left side to align with standard conic form:
\[ 4x^2 + y^2 = 4 \]
To obtain the standard ellipse form where the constant on the right is 1, divide every term in the equation by 4:
\[ \frac{4x^2}{4} + \frac{y^2}{4} = \frac{4}{4} \]
\[ \frac{x^2}{1} + \frac{y^2}{4} = 1 \]
Comparing this result to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can identify the semi-axis lengths:
\[ a^2 = 1 \implies a = 1 \]
\[ b^2 = 4 \implies b = 2 \]
The total area of the complete ellipse would be:
\[ \text{Total Area} = \pi \times a \times b = \pi \times 1 \times 2 = 2\pi \]
Recall that our original function \( y = 2\sqrt{1 - x^2} \) is only defined for \( y \geq 0 \).
The region bounded by this curve and the x-axis is the upper half of the ellipse.
Therefore, the required area is:
\[ \text{Required Area} = \frac{1}{2} \times \text{Total Area} \]
\[ \text{Required Area} = \frac{1}{2} \times 2\pi = \pi \]
The area is \( \pi \) square units.
Step 4: Final Answer:
By recognizing the curve as the upper half of an ellipse with semi-axes 1 and 2, we calculated the total area as \( 2\pi \) and correctly identified the bounded region's area as \( \pi \). This corresponds to Option (A).
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