Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y - 2.

OR


Using integration, find the area of the region enclosed between the two circles x2 + y2 = 4 and (x - 2)2 + y2 = 4.


Given curves are x2 = 4y and x = 4y – 2


Let A and B be the point of intersection of the given line and parabola


Coordinates of point A =


Coordinates of point B = (2, 1)


Area of region OBAO = Area of OBCO + Area of OACO


Firstly, we find the area of region OBCO










Now, we find the area of region OACO












Therefore, area of shaded region





OR


Given equations of circles are


x2 + y2 = 4 …(i)


and (x – 2)2 + y2 = 4 …(ii)


Consider the equation of circle x2 + y2 = 4


In this equation of circle, centre at the origin and radius is 2


Now, consider the equation (x – 2)2 + y2 = 4


In this equation of circle, centre is at (2, 0) and radius is 2


Solving eq. (i) and (ii), we get



⇒ x2 – (x – 2)2 = 0


⇒ x2 – (x2 + 4 – 4x) = 0


⇒ x2 – x2 – 4 + 4x = 0


⇒ 4x = 4


⇒ x = 1


Putting the value of x = 1 in eq. (i), we get


(1)2 + y2 = 4


⇒ y2 = 4 – 1


⇒ y = ±√3


Thus, the point of intersection of the given circles are A(1, √3) and B(1, -√3)



Required area (shaded region)


= Area of the region OACBO


= 2[Area of the region OACDO]


= 2[Area of the region OADO + Area of the region DCAD]






Now putting the upper and lower limits, we get









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