What is the area bounded by y= 16-x^2 , y ≥ 0 and the x -axis?
What is the area bounded by y=16−x2,y≥0 and the x-axis?
Options
A16π square units
B8π square units
C4π square units
D2π square units
Correct answer
B. 8π square units
Step-by-step solution
y=16−x2As, y≥0,16−x2≥0∴x∈[−4,4]Now,fromI=\int \sqrt{a^2-x^2} \cdot d x=\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1}(ax)+C
Here a2=16∴ Area bounded by the curve
I=∫−44(4)2−x2⋅dx=[2x16−x2]−44+216[sin−1(4x)]−44=0+8⋅[sin−1(44)−sin−1(−44)]=8[sin−1(1)+sin−1(1)]=16⋅sin−1(1)=16⋅2π=8π square units $
