The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2 sq.unitb...
The given curves are y² = 4ax [right parabola] .....(1)
And, x² = 4ay [upward parabola] .......(2)
On squaring both sides of equation 1, we get,
(y²)² = 16a²x²
y⁴ = 16a²(4ay) [From eq. 2]
y(y³ - 64a³) = 0
y = 0, y = 4a
When y = 0, x = 0
When y = 4a, x = 4a
Thus, the given parabolas intersect each other at O(0,0) and A(4a,4a). Then, the shaded part in the figure is the required area.
For the curve y² = 4ax
y = 2√a √x = f(x)
And, for the curve x² = 4ay,
y = x² / 4a = g(x)
∫₀⁴ᵃ [ f(x) - g(x) ] dx
= ∫₀⁴ᵃ 2√a √x dx - ∫₀⁴ᵃ (x² / 4a) dx
= (4/3) √a (4a)³/² - (1/12a) (4a)³
= (32a² / 3) - (16a² / 3)
= (16a² / 3) sq. units