Question

Using integration, find the area of the region {(x, y) : x² – 4y ≤ 0, y – x ≤ 0}.

Answer 1

Here, x² – 4y ≤ 0

x² ≤ 4y

This is a parabola.

y – x ≤ 0

y ≤ x

This is a straight line.

Finding point of intersection P:

Solving x² = 4y and y = x

x² = 4x

x² − 4x = 0

x(x − 4) = 0

So, x = 0, x = 4

For x = 0

y = x = 0

∴ O(0, 0)

For x = 4

y = x = 4

∴ P(4, 4)

Finding area to be shaded:

Now, our region is {(x, y) : x² – 4y ≤ 0, y – x ≤ 0}

Let’s take point (3, 1), which is below the parabola and line.

For parabola and point (3, 1):

x² – 4y ≤ 0

3² – 4(1) ≤ 0

9 – 4 ≤ 0

5 ≤ 0

This is not true.

So, (3, 1) will not be in the shaded region of the parabola.

Thus, x² – 4y ≤ 0 means the region above the parabola x² = 4y.

For line and point (3, 1):

y – x ≤ 0

1 – 3 ≤ 0

–2 ≤ 0

This is true.

Therefore, the point (3, 1) is located in the shaded region of the line.

Thus, x² – 4y ≤ 0 means the region below the line x = y.

Therefore, the combined shaded region will be above the parabola, below the line, and between the line and the parabola.

Find area:

Area required = Area OQPR

Thus, Area OQPR = Area ORPS − Area OQPS

Area ORPS = `int_0^4 y  dx`

Here, y is the equation of line QP.

y = x

∴ Area ORPS = `int_0^4 x  dx`

= `[(x^2)/2]_0^4`

= `[4^2/2 - 0^2/2]`

= `16/2 - 0`

= 8 sq. units

Area OQPS = `int_0^4 y  dx`

y → Equation of parabola.

x² = 4y

`x^2/4 = y`

y = `x^2/4`

∴ Area OQPS = `int_0^4 x^2/4 dx`

= `1/4 int_0^4 x^2 dx`

= `1/4 xx [x^3/3]_0^4`

= `1/4 xx [4^3/3 - 0^3/3]`

= `4^2/3`

= `16/3` sq. units

Thus, Area required = Area ORPS − Area OQPS

= `8 - 16/3`

= `(8 xx 3 - 16)/3`

= `(24 - 16)/3`

= `8/3` sq. units