# Using integration find the area of the region {(x, y) : x2+y2⩽ 2ax, y2⩾ ax, x, y ⩾ 0}. - Mathematics

Using integration find the area of the region {(x, y) : x2+y2 2ax, y2 ax, x, y  0}.

#### Solution

Given:

x2+y22ax, y2ax, x, y0

x2+y22ax0,  y2ax, x, y0

x2+y22ax+a2a20,  y2ax, x, y0

(xa)2+y2a2,  y2ax, x, y0

To find the points of intersection of the circle [(xa)2+y2=a2] and the parabola

[y2=ax],

we will substitute y2=ax in (xa)2+y2=a2.

(xa)2+ax=a2

x2+a22ax+ax=a2

x(xa)=0

x=0, a

Therefore, the points of intersection are (0, 0), (a, a) and (a, a).

Now,

Area of the shaded region= I

Area of I from x=0 to x=a

=[int_0^a(sqrt(a^2-(x-a^2)))dx-int_0^asqrt(axd)x]

Let xa=t for the first part of the integral  int_0^a(sqrt(a^2-(x-a^2)))dx

dx=dt

:.A_I=int_(-a)^0sqrt(a^2-t^2)dt-2sqrta/3|x^(3/2)|_0^a

=|t/2sqrt(a^2-t^2)+1/2a^2sin^(-1)

=0-(-(pia^2)/4)-(2a^2)/3

A_I=(pi/4-2/3)a^2

Area of the shaded region = (pi/4-2/3)a^2square units

Concept: Area Under Simple Curves
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