Find the volume of the solid generated by rotating

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

The region lies under $y=e^x$ for $x\le0$, bounded by $x=0$ and $y=0$; we compute both volumes and cross-check the second.

Understanding the region

Since $e^x>0$ always and $e^x\to0$ as $x\to-\infty$, the region spans $x\in(-\infty,0]$, with $y=1$ at $x=0$.

Part 1: Rotation about the $x$-axis

Disk method: $A(x)=\pi e^{2x}$.

\[V_x = \pi\int_{-\infty}^0 e^{2x}\,dx = \pi\left[\frac{e^{2x}}{2}\right]_{-\infty}^0 = \pi\left(\frac12-0\right) = \frac{\pi}{2}\]

\[\boxed{V_x = \frac{\pi}{2} \text{ cubic units}}\]

Part 2: Rotation about the $y$-axis

Step 1: Shell method

A shell at $x\le0$ has radius $|x|=-x$ and height $e^x$:

\[V_y = \int_{-\infty}^0 2\pi(-x)e^x\,dx = -2\pi\int_{-\infty}^0 xe^x\,dx\]

Step 2: Integrate by parts

\[\int xe^x\,dx = xe^x-e^x = e^x(x-1)+C\]

At $x=0$: $e^0(0-1)=-1$. As $x\to-\infty$: $e^x(x-1)\to0$ (exponential decay beats linear growth, confirmed by L'Hôpital).

\[\int_{-\infty}^0 xe^x\,dx = -1-0 = -1\]

Step 3: Calculate the volume

\[V_y = -2\pi(-1) = 2\pi\]

\[\boxed{V_y = 2\pi \text{ cubic units}}\]

Step 4: Cross-check via the washer (disk-in-$y$) method

With $x=\ln y$ for $y\in(0,1]$, radius $|x|=-\ln y$:

\[V_y = \pi\int_0^1(\ln y)^2\,dy = \pi\left[y(\ln y)^2-2y\ln y+2y\right]_0^1 = \pi(0-0+2) = 2\pi \quad\checkmark\]

Comparison

\[\frac{V_y}{V_x} = \frac{2\pi}{\pi/2} = 4\]

Summary

\[\boxed{V_x = \frac{\pi}{2}\approx1.571, \qquad V_y = 2\pi\approx6.283 \text{ cubic units}}\]

Key insight: Even though the region under $y=e^x$ extends infinitely far to the left, both volumes of revolution converge, and the $y$-axis volume ($2\pi$) comes out exactly four times the $x$-axis volume ($\pi/2$) — a clean numerical coincidence arising from the specific exponential decay rate.