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
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
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \)} and \( x = 2 \) is:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:
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