Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the curve $y^2 = x^4(1 + x)$ about the $x$-axis.

We identify a genuinely closed loop of the curve and compute its volume, while noting the remaining branch gives an infinite volume.

Step 1: Analyze the curve

\[y = \pm x^2\sqrt{1+x}\]

Real values require $1+x\ge0$, i.e. $x\ge-1$. At $x=-1$ and $x=0$: $y=0$. For $x\in(-1,0)$, the curve forms a closed loop (since $x^2\sqrt{1+x}$ is bounded there); for $x>0$, $y\to\infty$ as $x\to\infty$, an unbounded branch.

Step 2: Disk method on the loop

\[A(x) = \pi y^2 = \pi x^4(1+x)\]

Step 3: Integrate over the loop

\[V_{\text{loop}} = \pi\int_{-1}^0 (x^4+x^5)\,dx = \pi\left[\frac{x^5}{5}+\frac{x^6}{6}\right]_{-1}^0\]

At $x=0$: $0$. At $x=-1$: $-\dfrac15+\dfrac16 = \dfrac{-6+5}{30}=-\dfrac{1}{30}$.

\[V_{\text{loop}} = \pi\left[0-\left(-\frac{1}{30}\right)\right] = \frac{\pi}{30}\]

\[\boxed{V_{\text{loop}} = \frac{\pi}{30} \text{ cubic units}}\]

Step 4: The unbounded branch

For $x\ge0$: $V_{\text{unbounded}} = \pi\int_0^{\infty}(x^4+x^5)\,dx$, which diverges (the integrand grows without bound), so this branch generates infinite volume.

Summary

The closed loop (from $x=-1$ to $x=0$) generates:

\[\boxed{V_{\text{loop}} = \frac{\pi}{30} \approx 0.105 \text{ cubic units}}\]

while the branch extending to $x=\infty$ generates infinite volume.

Key insight: An implicit curve can combine a genuinely closed, finite-volume loop with an unbounded branch in the same equation; only the loop between the curve's two real zero-crossings ($x=-1,0$) produces a well-defined finite solid of revolution.