Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curve $y = 2xe^{-2x}$ and its asymptote about the asymptote.

We use the disk method and reduce the resulting integral to a standard Gamma-function value.

Step 1: Find the asymptote

As $x\to\infty$: $y=2xe^{-2x}\to0$ (exponential decay dominates), so the asymptote is $y=0$. For $x\ge0$, $y\ge0$ throughout.

Step 2: Locate the maximum (for context)

$\frac{dy}{dx}=2e^{-2x}(1-2x)=0 \implies x=\frac12$, giving $y_{\max}=\frac1e\approx0.368$.

Step 3: Disk method

\[A(x) = \pi(2xe^{-2x})^2 = 4\pi x^2e^{-4x}\]

Step 4: Set up and evaluate the integral

\[V = 4\pi\int_0^{\infty}x^2e^{-4x}\,dx\]

Let $u=4x$: $x=\frac u4$, $dx=\frac14du$:

\[\int_0^{\infty}x^2e^{-4x}\,dx = \frac{1}{64}\int_0^{\infty}u^2e^{-u}\,du = \frac{1}{64}\Gamma(3) = \frac{2}{64} = \frac{1}{32}\]

Step 5: Calculate the volume

\[V = 4\pi\cdot\frac{1}{32} = \frac{\pi}{8}\]

\[\boxed{V = \frac{\pi}{8} \text{ cubic units}}\]

Step 6: Verification of $\Gamma(3)=2$

\[\int_0^{\infty}u^2e^{-u}\,du = \left[-e^{-u}(u^2+2u+2)\right]_0^{\infty} = 0-(-2) = 2 \quad\checkmark\]

Step 7: Numerical value

\[V = \frac{\pi}{8} \approx 0.393 \text{ cubic units}\]

Summary

\[\boxed{V = \frac{\pi}{8} \approx 0.393 \text{ cubic units}}\]

Key insight: The curve rises from $0$, peaks at $x=\frac12$, then decays exponentially toward its asymptote; the disk-method integral is exactly a rescaled Gamma-function integral, so the volume reduces to the clean rational multiple of $\pi$ despite the exponential complexity.