Find the area bounded by the curves x² = 4ay and y =

Find the area bounded by the curves $x^2 = 4ay$ and $y = \frac{8a^3}{x^2 + 4a^2}$, where $a > 0$.

We will find the intersection points, then integrate the difference of the two curves using symmetry about the $y$-axis.

Step 1: Understand the curves

First curve (parabola): $x^2 = 4ay \implies y = \frac{x^2}{4a}$. This is a parabola opening upward with vertex at the origin.

Second curve: $y = \frac{8a^3}{x^2 + 4a^2}$. This is a bell-shaped curve symmetric about the $y$-axis. At $x=0$: $y=\frac{8a^3}{4a^2}=2a$. As $x\to\pm\infty$: $y\to0$.

Step 2: Find intersection points

\[\frac{x^2}{4a} = \frac{8a^3}{x^2 + 4a^2}\]

Cross-multiplying:

\[x^2(x^2 + 4a^2) = 32a^4 \implies x^4 + 4a^2x^2 - 32a^4 = 0\]

Let $u = x^2$:

\[u^2 + 4a^2u - 32a^4 = 0\]

\[u = \frac{-4a^2 \pm \sqrt{16a^4 + 128a^4}}{2} = \frac{-4a^2 \pm 12a^2}{2}\]

\[u = 4a^2 \quad \text{or} \quad u = -8a^2\]

Since $u = x^2 \geq 0$, we take $u = 4a^2$, so $x = \pm 2a$.

At $x = 2a$: $y = \frac{(2a)^2}{4a} = a$. By symmetry, at $x=-2a$: $y=a$ also. The curves intersect at $(-2a, a)$ and $(2a, a)$.

Step 3: Determine which curve is above

At $x = 0$: parabola gives $y=0$, second curve gives $y=2a$, so the second curve is above. Checking at $x=a$: parabola gives $y=\frac{a}{4}$, second curve gives $y=\frac{8a^3}{5a^2}=\frac{8a}{5}$, confirming the second curve remains above throughout $(-2a,2a)$.

Step 4: Set up the integral using symmetry

\[A = 2\int_0^{2a} \left[\frac{8a^3}{x^2 + 4a^2} - \frac{x^2}{4a}\right] dx\]

Step 5: Evaluate the first integral

\[\int_0^{2a} \frac{8a^3}{x^2 + 4a^2} \, dx = 8a^3 \cdot \frac{1}{2a}\left[\arctan\frac{x}{2a}\right]_0^{2a} = 4a^2\left[\arctan 1 - 0\right] = 4a^2\cdot\frac{\pi}{4} = \pi a^2\]

Step 6: Evaluate the second integral

\[\int_0^{2a} \frac{x^2}{4a} \, dx = \frac{1}{4a}\left[\frac{x^3}{3}\right]_0^{2a} = \frac{1}{4a}\cdot\frac{8a^3}{3} = \frac{2a^2}{3}\]

Step 7: Calculate the total area

\[A = 2\left[\pi a^2 - \frac{2a^2}{3}\right] = 2\pi a^2 - \frac{4a^2}{3}\]

\[\boxed{A = a^2\left(2\pi - \frac{4}{3}\right) = \frac{a^2(6\pi - 4)}{3} \text{ square units}}\]

Step 8: Numerical value

\[A \approx 4.950\,a^2 \text{ square units}\]

Summary

\[\boxed{A = \frac{a^2(6\pi-4)}{3} \approx 4.950a^2 \text{ square units}}\]

Key insight: The rational function creates a bell-shaped region above the parabola, with the area involving $\pi$ from the arctangent integral.