1. **Problem Statement:** Find the area bounded by the parabola $y^2 = 4ax$ and the parabola $x^2 = 4ay$ using double integration. 2. **Understanding the curves:** - The first parabola is $y^2 = 4ax$, which opens to the right. - The second parabola is $x^2 = 4ay$, which opens upwards. 3. **Find the points of intersection:** From $y^2 = 4ax$, express $x$ as $x = \frac{y^2}{4a}$. Substitute into $x^2 = 4ay$: $$\left(\frac{y^2}{4a}\right)^2 = 4ay \implies \frac{y^4}{16a^2} = 4ay \implies y^4 = 64 a^3 y.$$ Divide both sides by $y$ (considering $y \neq 0$): $$y^3 = 64 a^3 \implies y = 4a.$$ Also, $y=0$ is a solution. 4. **Corresponding $x$ values:** - For $y=0$, $x=0$. - For $y=4a$, $x = \frac{(4a)^2}{4a} = \frac{16 a^2}{4a} = 4a$. So the curves intersect at points $(0,0)$ and $(4a,4a)$. 5. **Set up the double integral:** The region is bounded between $y=0$ and $y=4a$. For each fixed $y$, $x$ varies between the parabola $x = \frac{y^2}{4a}$ (left) and $x = \sqrt{4ay} = 2\sqrt{ay}$ (right). 6. **Area $A$ is:** $$A = \int_0^{4a} \left(2\sqrt{ay} - \frac{y^2}{4a}\right) dy.$$ 7. **Evaluate the integral:** Split the integral: $$A = \int_0^{4a} 2\sqrt{ay} \, dy - \int_0^{4a} \frac{y^2}{4a} \, dy.$$ Calculate each: - $$\int_0^{4a} 2\sqrt{ay} \, dy = 2\sqrt{a} \int_0^{4a} y^{1/2} \, dy = 2\sqrt{a} \cdot \frac{2}{3} y^{3/2} \Big|_0^{4a} = \frac{4\sqrt{a}}{3} (4a)^{3/2}.$$ Note that $(4a)^{3/2} = (4a)^{1} \cdot (4a)^{1/2} = 4a \cdot 2\sqrt{a} = 8 a^{3/2}$. So, $$\frac{4\sqrt{a}}{3} \times 8 a^{3/2} = \frac{32 a^2}{3}.$$ - $$\int_0^{4a} \frac{y^2}{4a} \, dy = \frac{1}{4a} \int_0^{4a} y^2 \, dy = \frac{1}{4a} \cdot \frac{(4a)^3}{3} = \frac{1}{4a} \cdot \frac{64 a^3}{3} = \frac{16 a^2}{3}.$$ 8. **Subtract to find area:** $$A = \frac{32 a^2}{3} - \frac{16 a^2}{3} = \frac{16 a^2}{3}.$$ **Final answer:** $$\boxed{\frac{16 a^2}{3}}.$$