Question

Area of the Region bounded by the curve y=√49-x2 and x-axis is .

• (A) 49 π sq. units
• (B) 49 π/2 sq. units
• (C) 49 π/4 sq. units
• (D) 98 π sq. units

Answer 1

The Correct Option is B

The given curve is y = √(49 - x^2). 

This is the upper half of a circle with center at the origin and radius 7. 

The area bounded by this curve and the x-axis is the area of the upper half of the circle. 

The area of a circle with radius r is given by πr^2. 

Therefore, the area of the upper half of the circle with radius 7 is: (1/2)π(7^2) = 49π/2 square units. 

Hence, the correct option is (B) 49π/2 sq. units. 
 

Answer 2

To find the area of the region bounded by the curve \( y = \sqrt{49 - x^2} \) and the x-axis, we need to recognize that this curve represents the upper semicircle of a circle centered at the origin with radius 7 (since \(\sqrt{49 - x^2}\) is the equation of a circle of radius 7).
 Steps to solve:
1. Equation of the semicircle:
  \[ y = \sqrt{49 - x^2} \]This is the upper semicircle of a circle with radius \( 7 \) centered at the origin.
2. Limits of integration:
  The semicircle extends from \( x = -7 \) to \( x = 7 \).
3. Area of the semicircle:
  The area of a full circle with radius \( r \) is given by:
  \[ \text{Area}_{\text{circle}} = \pi r^2 \]
Since the radius \( r \) is 7:
  \[ \text{Area}_{\text{circle}} = \pi \times 7^2 = 49\pi \]
4. Area of the upper semicircle:
  The area of the upper semicircle is half of the area of the full circle:
  \[ \text{Area}_{\text{semicircle}} = \frac{1}{2} \times 49\pi = \frac{49\pi}{2} \]
 Final Result:
The area of the region bounded by the curve \( y = \sqrt{49 - x^2} \) and the x-axis is:
\[ \boxed{\frac{49\pi}{2} \text{ sq. units}} \]
Hence, the correct option is:
(B) 49 π/2 sq. units