Question

The area (in sq. units) of the region bounded by y = $2\sqrt{1-x^2}$, x $\in$ [0,1] and x-axis is equal to

• A. 1
• B. 2
• C. $\frac{\pi}{2}$
  (D) $\frac{\pi}{4}$
• D.

Hint:

Recognizing that the equation represents a standard geometric shape (like a circle or ellipse) can be much faster than performing the integration. Always try to identify the curve first before jumping into calculus.

Answer 1

The Correct Option is C

Step 1: Problem Interpretation:
The equation $y = 2\sqrt{1-x^2}$ defines a curve. Its area can be determined using either definite integration or geometric properties of an ellipse.
Step 2: Solution Approaches:
Method 1: Geometric Analysis
Transforming the equation $y = 2\sqrt{1-x^2}$ yields $\frac{y}{2} = \sqrt{1-x^2}$. Squaring both sides (for $y \ge 0$) results in $\frac{y^2}{4} = 1 - x^2$, which rearranges to $x^2 + \frac{y^2}{4} = 1$. This is the equation of an ellipse centered at the origin with semi-axes $a = 1$ and $b = 2$. The area of an ellipse is $\pi ab$. The condition $y \ge 0$ and $x \in [0,1]$ restricts the area to the first quadrant.
Method 2: Calculus Integration
The area is calculated via the definite integral $A = \int_{0}^{1} 2\sqrt{1-x^2} dx$.
Step 3: Detailed Calculations:
Method 1 (Geometric):
The total area of the ellipse $x^2 + \frac{y^2}{4} = 1$ is $\pi ab = \pi(1)(2) = 2\pi$. The specified region corresponds to one-quarter of this ellipse (first quadrant).
\[ A = \frac{1}{4} (2\pi) = \frac{\pi}{2} \]Method 2 (Integration):
\[ A = \int_{0}^{1} 2\sqrt{1-x^2} dx = 2 \int_{0}^{1} \sqrt{1-x^2} dx \]Using the antiderivative $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$ with $a=1$:
\[ A = 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{0}^{1} \]Evaluating the definite integral:
\[ A = 2 \left[ \left( \frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\sin^{-1}(1) \right) - \left( \frac{0}{2}\sqrt{1-0^2} + \frac{1}{2}\sin^{-1}(0) \right) \right] \]\[ A = 2 \left[ \left( 0 + \frac{1}{2} \cdot \frac{\pi}{2} \right) - (0 + 0) \right] \]\[ A = 2 \left[ \frac{\pi}{4} \right] = \frac{\pi}{2} \]Step 4: Conclusion:
The area of the specified region is $\frac{\pi}{2}$ square units.