Using integration find the area of the region {(x, y) : x2+y2⩽ 2ax, y2⩾ ax, x, y ⩾ 0}.
Solution
Given:
x2+y2≤2ax, y2≥ax, x, y≥0
⇒x2+y2−2ax≤0, y2≥ax, x, y≥0
⇒x2+y2−2ax+a2−a2≤0, y2≥ax, x, y≥0
⇒(x−a)2+y2≤a2, y2≥ax, x, y≥0
To find the points of intersection of the circle [(x−a)2+y2=a2] and the parabola
[y2=ax],
we will substitute y2=ax in (x−a)2+y2=a2.
(x−a)2+ax=a2
⇒x2+a2−2ax+ax=a2
⇒x(x−a)=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 x−a=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^2`square units