The area bounded by the parabola y^2 = 4ax and x^2 = 4ay is : Option 1: 8a^3/3 Option 2: 16a^2/3 Option 3: 32a^2/3 Option 4: 64a^2/3

Answer: 16a^2/3. Solution: The two parabolas intersect at (0,0) and (4a, 4a). Area = _0^4a ( √4ax - x^2/4a ) dx = _0^4a ( 2√a√x - x^2/4a ) dx = [ (4√a * x^(3/2))/3 - x^3/12a ]_0^4a = 32a^2/3 - 16a^2/3 = 16a^2/3. Note: Option 1 has incorrect units (a^3 instead of a^2); the intended answer matches 16a^2/3.

Calculus

Application of Integrals

Medium

The area bounded by the parabola $y^2 = 4ax$ and $x^2 = 4ay$ is :

$\frac{8a^3}{3}$

$\frac{16a^2}{3}$

$\frac{32a^2}{3}$

$\frac{64a^2}{3}$

✅ Correct Option: 2

The two parabolas intersect at $(0,0)$ and $(4a, 4a)$ .

Area $= \int_{0}^{4a} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) dx$

$= \int_{0}^{4a} \left( 2\sqrt{a}\sqrt{x} - \frac{x^2}{4a} \right) dx$

$= \left[ \frac{4\sqrt{a} \cdot x^{3/2}}{3} - \frac{x^3}{12a} \right]_{0}^{4a}$

$= \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3}$ .

Note: Option 1 has incorrect units ( $a^3$ instead of $a^2$ ); the intended answer matches $\frac{16a^2}{3}$ .

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The area bounded by the parabola $y^2 = 4ax$ and $x^2 = 4ay$ is :