Question

Find the area of the region bounded by the curve y² = 2y – x and 2y = 3x

Answer 1

Given curve is y² = 2y-x and line 2y = 3x.

For intersection point of line and curve, we have

y² = 2y - 2y/3

⇒ y² = 4y/3

⇒ y = 0 or y = 4/3

Hence, intersection points are (0, 0), (8/9, 4/3).

Given curve is y² = 2y-x

⇒ y² - 2y+1 = -x+1

⇒ (y-1)² = -(x-1)

Which is a parabola of vertex (1, 1) and axis of parabola is negative x-axis.

Since, x=0 gives y=0 and y=2

Hence, parabola touches y-axis at points (0, 0) and (0, 2).

Since, y=0 gives x=0.

Hence, parabola touches x-axis at (0, 0).

We observed that Region OABO is region bounded by given curve and line.

Where O(0, 0), A(1 ,1) and B(8/9, 4/3).

Region OABO is bounded between line x = 2y/3 and curve x = 2y-y²

And lines y=0 to y=4/3.

Required area = Area of region 

OABO = ₀∫^4/3(2y - y² - 2y/3)dy

= ₀∫^4/3 (4y/3 - y²) dy

\(=\frac{4}{3}(\frac{y^2}{2})_0^{4/3}-(\frac{y^3}{3})_0^{4/3}\)

= 4/3(16/18 - 0) - 1/3(64/27 - 0)

= 64/54 - 64/81 = 64/27(1/2 - 1/3)

= 64/27 x 3-2/6 = 64/27 x 1/6 = 32/81

Hence required bounded region is 32/81 square units.