Question

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

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( \pi/2 \)
• D. \( 4\pi \)

Hint:

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.

Answer 1

The Correct Option is A

Concept: Equations containing square roots of the form \[ y=b\sqrt{1-\frac{x^2}{a^2}} \] generally represent the upper half of an ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] Instead of integrating directly, recognizing the geometric figure often makes the problem much easier.

Step 1: Converting the equation into standard form.
Given: \[ y=2\sqrt{1-x^2} \] Squaring both sides: \[ y^2=4(1-x^2) \] Expand: \[ y^2=4-4x^2 \] Rearrange: \[ 4x^2+y^2=4 \] Divide by \(4\): \[ x^2+\frac{y^2}{4}=1 \] This is the equation of an ellipse. Comparing with: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] we get: \[ a^2=1,\qquad b^2=4 \] Thus, \[ a=1,\qquad b=2 \]

Step 2: Understanding the region represented.
The given equation is: \[ y=2\sqrt{1-x^2} \] Since square root always gives non-negative values, \[ y\ge0 \] Therefore the curve represents only the upper half of the ellipse.

Step 3: Finding the total area of the ellipse.
Area of ellipse: \[ \text{Area}=\pi ab \] Substitute values: \[ \text{Area}=\pi(1)(2)=2\pi \]

Step 4: Finding the required bounded area.
Since only upper half is considered, \[ \text{Required Area}=\frac12(2\pi) \] Thus, \[ \text{Required Area}=\pi \] Hence, \[ \boxed{\pi} \]