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: Understanding the Concept:
The given curve $y = 2\sqrt{1-x^2}$ represents a portion of an ellipse. We can find the area either by setting up a definite integral or by using the geometric formula for the area of an ellipse.
Step 2: Key Formula or Approach:
Method 1: Geometric Interpretation
Rearrange the equation of the curve:
\[ y = 2\sqrt{1-x^2} \implies \frac{y}{2} = \sqrt{1-x^2} \] Square both sides (assuming y $\ge$ 0):
\[ \left(\frac{y}{2}\right)^2 = 1 - x^2 \implies x^2 + \frac{y^2}{4} = 1 \] This is the standard equation of an ellipse centered at the origin with semi-minor axis a = 1 and semi-major axis b = 2. The area of a full ellipse is $\pi ab$. The given constraints are $y \ge 0$ (from the positive square root) and $x \in [0,1]$, which means we are finding the area of the ellipse in the first quadrant.
Method 2: Integration
The area is given by the integral $A = \int_{0}^{1} 2\sqrt{1-x^2} dx$.
Step 3: Detailed Explanation:
Using Method 1 (Geometry):
The total area of the ellipse $x^2 + \frac{y^2}{4} = 1$ is $\pi ab = \pi(1)(2) = 2\pi$.
The required area is the portion in the first quadrant (where x is from 0 to 1 and y is positive). This is exactly one-quarter of the total area of the ellipse.
\[ A = \frac{1}{4} (\text{Total Area}) = \frac{1}{4} (2\pi) = \frac{\pi}{2} \] Using Method 2 (Integration):
\[ A = \int_{0}^{1} 2\sqrt{1-x^2} dx = 2 \int_{0}^{1} \sqrt{1-x^2} dx \] We use the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$. Here, a = 1.
\[ A = 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{0}^{1} \] \[ A = 2 \left[ \left( \frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\sin^{-1}(1) \right) - \left( \frac{0}{2}\sqrt{1-0^2} + \frac{1}{2}\sin^{-1}(0) \right) \right] \] \[ A = 2 \left[ \left( 0 + \frac{1}{2} \cdot \frac{\pi}{2} \right) - (0 + 0) \right] \] \[ A = 2 \left[ \frac{\pi}{4} \right] = \frac{\pi}{2} \] Step 4: Final Answer:
The area of the region is $\frac{\pi{2}$} sq. units.