Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = x^2(1 - x^2)$ and $y = 0$ about the $x$-axis.

We identify the bounded region where $y\ge0$, then integrate the squared quartic using symmetry.

Step 1: Analyze the curve

\[y = x^2(1-x^2) = x^2-x^4\]

$y=0$ at $x=0,\pm1$; $y>0$ for $|x| < 1$ (excluding $0$) and $y<0$ for $|x|>1$. The function is even.

Maximum: $\frac{dy}{dx}=2x-4x^3=2x(1-2x^2)=0 \implies x=0,\pm\frac{1}{\sqrt2}$; at $x=\frac{1}{\sqrt2}$, $y=\frac12\left(1-\frac12\right)=\frac14$.

Step 2: Identify the bounded region

Rotating about the $x$-axis with $y=0$ as the other boundary uses the region where $y\ge0$, i.e. $-1\le x\le1$.

Step 3: Disk method

\[A(x) = \pi(x^2-x^4)^2 = \pi(x^4-2x^6+x^8)\]

Step 4: Integrate using symmetry

\[V = 2\int_0^1\pi(x^4-2x^6+x^8)\,dx = 2\pi\left[\frac{x^5}{5}-\frac{2x^7}{7}+\frac{x^9}{9}\right]_0^1\]

At $x=1$ (common denominator 315): $\dfrac{63}{315}-\dfrac{90}{315}+\dfrac{35}{315}=\dfrac{8}{315}$.

\[V = 2\pi\cdot\frac{8}{315} = \frac{16\pi}{315}\]

\[\boxed{V = \frac{16\pi}{315} \text{ cubic units}}\]

Step 5: Numerical value

\[V = \frac{16\pi}{315} \approx 0.159 \text{ cubic units}\]

Summary

\[\boxed{V = \frac{16\pi}{315} \approx 0.159 \text{ cubic units}}\]

Key insight: The curve $y=x^2(1-x^2)$ forms two symmetric humps meeting at the origin with height only $\frac14$, so despite the region's visible width the resulting solid of revolution is quite slender, giving a small compact volume.