Q102 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
If the area under the curve $y=\sqrt{a^2-x^2}$ included between the lines $x=0$ and $x=a$ is 4 sq units. Then the value of ' $a$ ' is
- A$\dfrac{2}{\sqrt{\pi}}$
- B$\dfrac{4}{\pi}$
- C$\dfrac{16}{\sqrt{\pi}}$
- Correct answerD$\dfrac{4}{\sqrt{\pi}}$
Solution
The area under the curve $y = \sqrt{a^2 - x^2}$ from $x = 0$ to $x = a$ is given by the integral $A = \int_{0}^{a} \sqrt{a^2 - x^2} dx$.
Using the standard integral formula $\int \sqrt{a^2 - x^2} dx = \dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left( \dfrac{x}{a} \right)$, we evaluate the definite integral:
$A = \left[ \dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left( \dfrac{x}{a} \right) \right]_{0}^{a}$
Substituting the limits:
$A = \left( \dfrac{a}{2} \sqrt{a^2 - a^2} + \dfrac{a^2}{2} \sin^{-1} (1) \right) - \left( 0 + \dfrac{a^2}{2} \sin^{-1} (0) \right)$
$A = 0 + \dfrac{a^2}{2} \left( \dfrac{\pi}{2} \right) - 0 = \dfrac{\pi a^2}{4}$
Given that the area $A = 4$, we have $\dfrac{\pi a^2}{4} = 4$.
Solving for $a^2$: $a^2 = \dfrac{16}{\pi}$.
Therefore, $a = \sqrt{\dfrac{16}{\pi}} = \dfrac{4}{\sqrt{\pi}}$.
Answer: $\dfrac{4}{\sqrt{\pi}}$
Using the standard integral formula $\int \sqrt{a^2 - x^2} dx = \dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left( \dfrac{x}{a} \right)$, we evaluate the definite integral:
$A = \left[ \dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left( \dfrac{x}{a} \right) \right]_{0}^{a}$
Substituting the limits:
$A = \left( \dfrac{a}{2} \sqrt{a^2 - a^2} + \dfrac{a^2}{2} \sin^{-1} (1) \right) - \left( 0 + \dfrac{a^2}{2} \sin^{-1} (0) \right)$
$A = 0 + \dfrac{a^2}{2} \left( \dfrac{\pi}{2} \right) - 0 = \dfrac{\pi a^2}{4}$
Given that the area $A = 4$, we have $\dfrac{\pi a^2}{4} = 4$.
Solving for $a^2$: $a^2 = \dfrac{16}{\pi}$.
Therefore, $a = \sqrt{\dfrac{16}{\pi}} = \dfrac{4}{\sqrt{\pi}}$.
Answer: $\dfrac{4}{\sqrt{\pi}}$
Hence, the correct answer is option: 'D'.