Find the area bounded by the arc of the curve 81y² =

Find the area bounded by the arc of the curve $81y^2 = 4x^2(9 - x^2)$.

We will use symmetry about both axes to reduce the enclosed loop to a single elementary integral.

Step 1: Simplify the curve equation

\[y^2 = \frac{4x^2(9-x^2)}{81} \implies y = \pm\frac{2x\sqrt{9-x^2}}{9}\]

Step 2: Determine the domain

We need $9-x^2\geq0$, i.e. $x\in[-3,3]$.

Step 3: Find key points

At $x=0$: $y=0$. At $x=\pm3$: $y=0$. So the curve passes through $(-3,0)$, $(0,0)$, $(3,0)$.

Step 4: Find the maximum height

Differentiating $y=\frac29x\sqrt{9-x^2}$:

\[\frac{dy}{dx} = \frac29\left[\sqrt{9-x^2} - \frac{x^2}{\sqrt{9-x^2}}\right] = \frac{2(9-2x^2)}{9\sqrt{9-x^2}}\]

Setting this to zero: $x^2=\frac92\implies x=\pm\frac{3\sqrt2}{2}$. At $x=\frac{3\sqrt2}{2}$: $y=1$. So the maximum height of the upper branch is $1$.

Step 5: Set up the integral using symmetry

By symmetry about both axes:

\[A = 4\int_0^3 \frac{2x\sqrt{9-x^2}}{9}\,dx = \frac89\int_0^3 x\sqrt{9-x^2}\,dx\]

Step 6: Evaluate the integral

Let $u=9-x^2$, $du=-2x\,dx$; $x=0\Rightarrow u=9$, $x=3\Rightarrow u=0$:

\[\int_0^3 x\sqrt{9-x^2}\,dx = \frac12\int_0^9 u^{1/2}\,du = \frac13\left[u^{3/2}\right]_0^9 = \frac13\cdot27 = 9\]

Step 7: Calculate the total area

\[A = \frac89\cdot9 = 8\]

\[\boxed{A = 8 \text{ square units}}\]

Summary

\[\boxed{A = 8 \text{ square units}}\]

Key insight: Despite the complicated-looking equation, the enclosed loop — symmetric about both axes, passing through $(\pm3,0)$ and $(0,0)$, with maximum height $1$ — has the simple integer area $8$.