AOB is the positive quadrant of the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1` in which `OA=a,OB=b`, The area between the arc AB and chord AB of the ellipse is

Options

A. `1/2ab(pi+2)`
B. `1/4ab(x-4)`
C. `1/4ab(pi-2)`
D. None

Answer

To find the area between the arc AB and the chord AB of the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) in the positive quadrant, we can follow these steps:

### Step 1: Understand the Geometry
We have an ellipse in the first quadrant with points A and B defined as follows:
- Point A is at (a, 0)
- Point B is at (0, b)
- The area we want to find is between the arc AB of the ellipse and the chord AB.

### Step 2: Area of Triangle AOB
First, we calculate the area of triangle AOB formed by points O(0,0), A(a,0), and B(0,b). The area \(A_{triangle}\) of triangle AOB can be calculated using the formula:
\[ A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times b = \frac{ab}{2} \]

### Step 3: Area Under the Ellipse Arc
Next, we need to find the area under the arc AB of the ellipse from x = 0 to x = a. To do this, we express y in terms of x using the ellipse equation:
\[ y = b \sqrt{1 - \frac{x^2}{a^2}} \]
Now, we can set up the integral to find the area under the curve:
\[ A_{arc} = \int_0^a b \sqrt{1 - \frac{x^2}{a^2}} \, dx \]

### Step 4: Solve the Integral
To solve the integral, we can use the substitution \( x = a \sin(\theta) \), which gives \( dx = a \cos(\theta) d\theta \). The limits change from \( x = 0 \) (when \( \theta = 0 \)) to \( x = a \) (when \( \theta = \frac{\pi}{2} \)). The integral becomes:
\[ A_{arc} = \int_0^{\frac{\pi}{2}} b \sqrt{1 - \sin^2(\theta)} \cdot a \cos(\theta) \, d\theta = \int_0^{\frac{\pi}{2}} b a \cos^2(\theta) \, d\theta \]
Using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \), we can rewrite the integral:
\[ A_{arc} = \frac{ab}{2} \int_0^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta \]
Calculating this integral:
\[ = \frac{ab}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_0^{\frac{\pi}{2}} = \frac{ab}{2} \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = \frac{ab\pi}{4} \]

### Step 5: Area Between Arc and Chord
Finally, the area between the arc AB and the chord AB is given by:
\[ A_{between} = A_{arc} - A_{triangle} = \frac{ab\pi}{4} - \frac{ab}{2} \]
To combine the terms:
\[ = \frac{ab\pi}{4} - \frac{2ab}{4} = \frac{ab(\pi - 2)}{4} \]

### Final Answer
Thus, the area between the arc AB and the chord AB of the ellipse is:
\[ \frac{ab(\pi - 2)}{4} \]