Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = e^{3x} + e^x$, $x = 0$, and $y = e^3 + e$ about the $y$-axis.

We use the shell method, splitting the integral into two pieces each requiring integration by parts.

Step 1: Understand the region

At $x=0$: $y=1+1=2$. At $x=1$: $y=e^3+e$, matching the given upper bound — so the region runs from $x=0$ to $x=1$, bounded on the right by the curve and capped at $y=e^3+e$.

Step 2: Shell method

\[V = 2\pi\int_0^1 x(e^{3x}+e^x)\,dx = 2\pi\left[\int_0^1xe^{3x}\,dx+\int_0^1xe^x\,dx\right]\]

Step 3: Evaluate $\int_0^1xe^{3x}\,dx$

By parts ($u=x$, $dv=e^{3x}dx$): $\int xe^{3x}\,dx = \dfrac{e^{3x}}{9}(3x-1)+C$.

\[\left[\frac{e^{3x}}{9}(3x-1)\right]_0^1 = \frac{e^3}{9}(2)-\frac19(-1) = \frac{2e^3+1}{9}\]

Step 4: Evaluate $\int_0^1xe^x\,dx$

By parts: $\int xe^x\,dx = e^x(x-1)+C$.

\[[e^x(x-1)]_0^1 = e(0)-1(-1) = 1\]

Step 5: Calculate the volume

\[V = 2\pi\left(\frac{2e^3+1}{9}+1\right) = 2\pi\cdot\frac{2e^3+10}{9} = \frac{4\pi(e^3+5)}{9}\]

\[\boxed{V = \frac{4\pi(e^3+5)}{9} \text{ cubic units}}\]

Step 6: Numerical value

$e^3\approx20.086$, so $V\approx\dfrac{4\pi(25.086)}{9}\approx35.0$ cubic units.

Summary

\[\boxed{V = \frac{4\pi(e^3+5)}{9} \approx 35.0 \text{ cubic units}}\]

Key insight: Because $y=e^{3x}+e^x$ is a sum of two exponentials with different rates, the shell-method integral splits naturally into two separate integration-by-parts computations, each contributing its own piece to the final combined result involving $e^3$.