Smaller area enclosed by the circle x2+y2=4 and the line x+y=2 is
Smaller area enclosed by the circle x2+y2=4 and the line x+y=2 is
2(π-2)
π-2
2π-1
2(π+2)
The Correct Option is B
Solution and Explanation
The smaller area enclosed by the circle,x2+y2=4,and the line,x+y=2,is
represented by the shaded area ACBA as
It can be observed that,
Area ACBA=Area OACBO-Area(ΔOAB)=
\[\int_{1}^{2} \sqrt{4-x^2} \,dx\]\[-\int_{0}^{2} (2-x) \,dx\]=[\(\frac{x}{2}\)\(\sqrt{4-x^2}\)+\(\frac{4}{2}\)sin-1\(\frac{x}{2}\)]20-[2x-\(\frac{x^2}{2}\)]20
=[2.π/2]-[4-2]
=(π-2)units.
Thus,the correct answer is B.
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Concepts Used:
Area between Two Curves
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.
Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,
