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