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

Question

Prepp Team · Accepted Answer

Area Calculation: Bounded Region

The question asks for the area of the region enclosed by the curve $y = 2\sqrt{1-x^2}$, the x-axis, and the vertical lines $x=0$ and $x=1$. We need to determine this area in square units.

Curve $y = 2\sqrt{1-x^2}$ Analysis

Let's examine the equation $y = 2\sqrt{1-x^2}$:

The presence of $\sqrt{1-x^2}$ means that $1-x^2 \ge 0$, which implies $x$ must be in the interval $[-1, 1]$.
Also, the square root symbol denotes the principal (non-negative) root, so $y \ge 0$. This indicates we are dealing with the upper part of a curve.
Squaring both sides gives $y^2 = (2\sqrt{1-x^2})^2$, which simplifies to $y^2 = 4(1-x^2)$.
Rearranging terms, we get $y^2 = 4 - 4x^2$, leading to $4x^2 + y^2 = 4$.
Dividing the entire equation by 4 yields $\frac{x^2}{1} + \frac{y^2}{4} = 1$.
This is the standard form of an ellipse centered at the origin $(0,0)$ with semi-axes $a=1$ (along the x-axis) and $b=2$ (along the y-axis).
Given $y \ge 0$, the equation $y = 2\sqrt{1-x^2}$ specifically represents the upper half of this ellipse.

Region Bounded by Curve and x-axis

The boundaries defining the area are:

The upper semi-ellipse given by $y = 2\sqrt{1-x^2}$.
The x-axis, which is the line $y=0$.
The specified interval for $x$ is $[0, 1]$, corresponding to the vertical lines $x=0$ and $x=1$.

This describes the area under the curve $y = 2\sqrt{1-x^2}$ from $x=0$ to $x=1$. Geometrically, this is the portion of the ellipse located in the first quadrant.

Area Calculation: Geometric Method

The standard equation for the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. For our ellipse, $a=1$ and $b=2$. The area of a full ellipse is calculated using the formula $A_{ellipse} = \pi ab$.

For this ellipse, the total area is $A_{full} = \pi \times 1 \times 2 = 2\pi$ square units.

The specified region covers the part of the ellipse in the first quadrant, which is exactly one-quarter of the total ellipse.

Therefore, the area of the region is:

Area $= \frac{1}{4} \times A_{full} = \frac{1}{4} \times 2\pi = \frac{2\pi}{4} = \frac{\pi}{2}$ square units.

Area Calculation: Integration Method

The area under a curve $y = f(x)$ from $x=a$ to $x=b$ can be found using the definite integral $A = \int_{a}^{b} y \, dx$.

In this case, the area integral is:

$ \text{Area} = \int_{0}^{1} 2\sqrt{1-x^2} \, dx $

We can solve this integral using trigonometric substitution. Let $x = \sin\theta$. This implies $dx = \cos\theta \, d\theta$. The limits of integration need to be adjusted accordingly:

When $x=0$, $\sin\theta = 0$, so $\theta = 0$.
When $x=1$, $\sin\theta = 1$, so $\theta = \frac{\pi}{2}$.

Substitute these into the integral:

$ \text{Area} = \int_{0}^{\pi/2} 2\sqrt{1-\sin^2\theta} (\cos\theta \, d\theta) $

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we have $\sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta}$. Since $0 \le \theta \le \frac{\pi}{2}$, $\cos\theta \ge 0$, thus $\sqrt{\cos^2\theta} = \cos\theta$.

$ \text{Area} = \int_{0}^{\pi/2} 2(\cos\theta)(\cos\theta \, d\theta) $ $ \text{Area} = \int_{0}^{\pi/2} 2\cos^2\theta \, d\theta $

Employ the double-angle identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$ \text{Area} = \int_{0}^{\pi/2} 2 \left( \frac{1 + \cos(2\theta)}{2} \right) \, d\theta $ $ \text{Area} = \int_{0}^{\pi/2} (1 + \cos(2\theta)) \, d\theta $

Now, perform the integration:

$ \text{Area} = \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{0}^{\pi/2} $

Evaluate the result at the limits:

$ \text{Area} = \left( \frac{\pi}{2} + \frac{1}{2}\sin\left(2 \times \frac{\pi}{2}\right) \right) - \left( 0 + \frac{1}{2}\sin(2 \times 0) \right) $ $ \text{Area} = \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) \right) - \left( 0 + \frac{1}{2}\sin(0) \right) $

Since $\sin(\pi) = 0$ and $\sin(0) = 0$:

$ \text{Area} = \left( \frac{\pi}{2} + 0 \right) - (0 + 0) $ $ \text{Area} = \frac{\pi}{2} $

Both methods confirm the area is $\frac{\pi}{2}$ square units.

Final Area Result

The area calculation, using either geometric properties of the ellipse or definite integration, shows that the region bounded by $y = 2\sqrt{1-x^2}$ and the x-axis between $x=0$ and $x=1$ is $\frac{\pi}{2}$ square units.