Problem Statement
Find the area of the region enclosed between the two parabolas $y^2 = 4ax$ and $x^2 = 4ay$ ($a > 0$) using definite integration.
Verified Solution & Marking Scheme
Find Points of Intersection
Given parabolas: (1) $y^2 = 4ax$, (2) $x^2 = 4ay \implies y = \frac{x^2}{4a}$.
Substituting $y$ into (1):
$\left( \frac{x^2}{4a} \right)^2 = 4ax \implies \frac{x^4}{16a^2} = 4ax \implies x^4 - 64a^3 x = 0$
$x(x^3 - 64a^3) = 0 \implies x = 0 \text{ or } x = 4a$
When $x = 0$, $y = 0 \implies (0, 0)$.
When $x = 4a$, $y = \frac{(4a)^2}{4a} = 4a \implies (4a, 4a)$.
Set up the Definite Integral for Area
For $x \in [0, 4a]$, the upper curve is $y_1 = \sqrt{4ax} = 2\sqrt{a}\sqrt{x}$ and the lower curve is $y_2 = \frac{x^2}{4a}$.
$A = \int_{0}^{4a} (y_1 - y_2) \, dx = \int_{0}^{4a} \left( 2\sqrt{a} x^{1/2} - \frac{x^2}{4a} \right) \, dx$
Evaluate the Integrals
$A = \left[ 2\sqrt{a} \frac{x^{3/2}}{3/2} - \frac{x^3}{12a} \right]_{0}^{4a} = \left[ \frac{4\sqrt{a}}{3} x^{3/2} - \frac{x^3}{12a} \right]_{0}^{4a}$
At $x = 4a$:
$x^{3/2} = (4a)^{3/2} = 8a^{3/2}$
$\frac{4\sqrt{a}}{3} (8a^{3/2}) = \frac{32a^2}{3}$
$\frac{(4a)^3}{12a} = \frac{64a^3}{12a} = \frac{16a^2}{3}$
$A = \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3}$ sq. units.