Draw a rough sketch and find the area of the region bounded by the parabolas y2=4x and x2=4y, using the method of integration.
Given the boundaries of the area to be found are,
• the first parabola, y2 = 4x ---(1)
• the second parabola, x2 = 4y ---- (2)
From the equation, of the first parabola, y2 = 4x
• the vertex at (0,0) i.e. the origin
• Symmetric about the x-axis, as it has the even power of y.
From the equation, of the second parabola, x2 = 4y
• the vertex at (0,0) i.e. the origin
• Symmetric about the y-axis, as it has the even power of x.
Now to find the point of intersection of (1) and (2), substitute in (1)
x4 = 64x
x(x3- 64) = 0
x = 0 (or) x = 4
Substituting x in (2), we get y = 0 (or) y = 4
So the two points, A and B where (1) and (2) meet are A = (4,4) and O= (0,0)
Consider the first parabola, y2 = 4x, can be re-written as
----- (3)
Consider the parabola, x2 = 4y, can be re-written as
----- (4)
Let us drop a perpendicular from A on to x-axis. The base of the perpendicular is D = (4, 0)
Now, the area to be found will be the area is
Area of the required region = Area of OBACO.
Area of OBACO= Area of OBADO- Area of OCADO
Area of OBACOis
[Using the formula, ]
The Required Area of OBACOsq. units