The regions bounded by the curve y = e^-xsqrt(sin x) (x
The regions bounded by the curve $y = e^{-x}\sqrt{\sin x}$ $(x\ge0)$ and the $x$-axis are rotated about the $x$-axis. Find the sum of the volumes of the solids generated by this rotation.
Explanation of Method
Since $\sqrt{\sin x}$ is only real where $\sin x\ge0$, the curve exists on infinitely many disjoint intervals $[2k\pi,(2k+1)\pi]$, each producing its own solid of revolution when rotated about the $x$-axis. Computing the volume on one representative interval via the disk method, then exploiting the fact that each subsequent interval is the previous one shifted by $2\pi$ (with the integrand scaled by a factor of $e^{-4\pi}$ due to the exponential decay), turns the infinite collection of volumes into a geometric series that can be summed exactly.
- Identify the intervals where the curve is defined.
- Compute the volume on one interval using the disk method and integration by parts.
- Recognize that successive volumes form a geometric series and sum it.
Step 1 · Identify the intervals
The function is defined only when $\sin x\ge0$, which for $x\ge0$ occurs on $[2k\pi,(2k+1)\pi]$ for $k=0,1,2,\ldots$. Rotation about the $x$-axis on each such interval generates one solid, whose volume must be found and then summed over all $k$.
Step 2 · Volume formula and the integral to evaluate
The volume of revolution on $[a,b]$ is $V=\pi\int_a^b y^2\,dx$, and here $y^2 = e^{-2x}\sin x$. On $[2k\pi,(2k+1)\pi]$:
$$ V_k = \pi\int_{2k\pi}^{(2k+1)\pi} e^{-2x}\sin x\,dx. $$
Integrating by parts twice (a standard reduction for $\int e^{ax}\sin(bx)\,dx$) gives the antiderivative:
$$ \int e^{-2x}\sin x\,dx = -\frac15 e^{-2x}(2\sin x+\cos x) + C. $$
Step 3 · Evaluate on one interval and sum
At $x=(2k+1)\pi$: $\sin=0,\cos=-1$, giving $\tfrac15 e^{-2(2k+1)\pi}$. At $x=2k\pi$: $\sin=0,\cos=1$, giving $-\tfrac15 e^{-4k\pi}$. Subtracting:
$$ \int_{2k\pi}^{(2k+1)\pi} e^{-2x}\sin x\,dx = \frac15 e^{-4k\pi}\left(1+e^{-2\pi}\right), $$
so
$$ V_k = \frac{\pi(1+e^{-2\pi})}{5}\, e^{-4k\pi}. $$
Summing over $k=0,1,2,\ldots$, this is a geometric series with ratio $r=e^{-4\pi}<1$:
$$ V = \sum_{k=0}^{\infty} V_k = \frac{\pi(1+e^{-2\pi})}{5}\sum_{k=0}^{\infty}\left(e^{-4\pi}\right)^k = \frac{\pi(1+e^{-2\pi})}{5}\cdot\frac{1}{1-e^{-4\pi}}. $$
Summary Table
| Quantity | Value |
|---|---|
| Volume on interval $k$: $V_k$ | $\dfrac{\pi(1+e^{-2\pi})}{5}\,e^{-4k\pi}$ |
| Ratio of the geometric series | $e^{-4\pi}$ |
| Total volume $V$ | $\dfrac{\pi(1+e^{-2\pi})}{5(1-e^{-4\pi})}$ |
Additional Notes
- Because $e^{-4\pi}\approx3.5\times10^{-6}$, the volumes shrink extremely fast; the very first solid (from $[0,\pi]$) already accounts for well over $99.999\%$ of the total volume $V$.
- The equivalent factored form $V=\dfrac{\pi(e^{4\pi}+e^{2\pi})}{5(e^{4\pi}-1)}$ is obtained by multiplying numerator and denominator by $e^{4\pi}$; both forms are exact, but the $e^{-4k\pi}$ version makes the rapid geometric decay of successive lobes more transparent.
- This is a good example of how integration by parts (for one bounded piece) and the geometric series formula (for the infinite repetition) combine when a periodic-in-shape but decaying-in-amplitude curve generates infinitely many separate solids.
Result
The total volume of all solids of revolution is: