Question

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.

Answer 1

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 y = \(\frac{x^2}{4}\) in (1) \(\Big(\frac{x^2}{4}\Big)^2\) = 4x

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 

y = 2\(\sqrt{x}\) ........ (3)

Consider the parabola, x² = 4y, can be re-written as

y = \(\frac{x^2}{4}\) ....... (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 OBACO is

The Required Area of OBACO = \(\Big(\frac{16}{3}\Big)\)sq. units