Question

The area of the region \[ \{(x,y): \sin x \le y \le \sqrt{\pi^2-x^2}\} \] is:

• A. \(\pi^3\)
• B. \(\dfrac{\pi^3}{2}\)
• C. \(\dfrac{\pi^3}{4}\)
• D. \(\dfrac{\pi^3}{8}\)

Hint:

Whenever limits are symmetric and the integrand contains odd functions, check if the integral evaluates to zero to simplify calculations.

Answer 1

The Correct Option is B

Concept:
The area between two curves \(y=f(x)\) and \(y=g(x)\) is \(\displaystyle \int (g(x)-f(x))\,dx\).
The curve \(y=\sqrt{\pi^2-x^2}\) represents the upper semicircle of radius \(\pi\).
The function \(\sin x\) is odd, hence its integral over a symmetric interval about the origin is zero.
Step 1: Determine the limits of integration For \(y=\sqrt{\pi^2-x^2}\), we must have: \[ \pi^2-x^2 \ge 0 \Rightarrow -\pi \le x \le \pi \]
Step 2: Set up the area integral \[ \text{Area}=\int_{-\pi}^{\pi}\left(\sqrt{\pi^2-x^2}-\sin x\right)\,dx \]
Step 3: Split the integral \[ \text{Area}=\int_{-\pi}^{\pi}\sqrt{\pi^2-x^2}\,dx -\int_{-\pi}^{\pi}\sin x\,dx \]
Step 4: Evaluate the integrals Since \(\sin x\) is an odd function, \[ \int_{-\pi}^{\pi}\sin x\,dx=0 \] The remaining integral represents the area of a semicircle of radius \(\pi\): \[ \int_{-\pi}^{\pi}\sqrt{\pi^2-x^2}\,dx=\frac{1}{2}\pi(\pi)^2=\frac{\pi^3}{2} \]
Step 5: Final answer \[ \text{Area}=\frac{\pi^3}{2} \]