Find the area of the region included between the parabola y2 =3x and the circle x2+y2-6x=0, lying in the first quadrant.
Given the boundaries of the area to be found are,
• the circle, x2 + y2 – 6x = 0 ---(1)
• the parabola, y2 = 3x ---- (2)
• Area under 1st quadrant.
From the equation, of the first circle, x2 + y2 – 6x = 0
x2 – 6x + 9 + y2 – 9 = 0
(x-3)2 + y2 = 9
• the vertex at (3,0)
• the radius is 3 units.
From the equation, of the parabola , y2 = 3x
• the vertex at (0,0) i.e. the origin
• Symmetric about the x-axis, as it has the even power of y.
Now to find the point of intersection of (1) and (2), substitute y2 = 3x in (1)
x2 + 3x – 6x = 0
x2 – 3x = 0
x(x-3) = 0
x = 3 (or) x = 0
Substituting x in (2), we get y = ± 3 or y = 0
So the three points, A, B and •where (1) and (2) meet are A = (3,3) , B = (3,-3) and O=(0,0)
Consider the circle, x2 + y2 – 6x , can be re-written as
----- (3)
Consider the parabola, y2 = 3x, can be re-written as
----- (4)
Let us drop a perpendicular from A on to x-axis. The base of the perpendicular is C = (3, 0)
Now, the area to be found will be the area is
Area of the required region = Area between the circle and the parabola at OA.
Area of OA= Area under circle OAC - Area under parabola OAC
Area of OA is
[Using the formula, and ]
[sin-1(-1) = -90°]
Area of the required region is sq. units.