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.


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