1. **State the problem:** We need to find the value of $k$ (either 1 or 2) such that the volume of the solid formed by rotating the region bounded by the curve $y = e^x - k$, the x-axis, and the line $x = \ln 3$ about the x-axis is $\pi \ln 3$. 2. **Formula for volume of revolution about x-axis:** $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $f(x)$ is the function being rotated, and $a$, $b$ are the bounds. 3. **Set up the integral:** Here, $f(x) = e^x - k$, $a = 0$ (since the region is bounded by the x-axis, the lower bound is where $y=0$), and $b = \ln 3$. 4. **Check the lower bound:** Solve $e^x - k = 0$ for $x$: $$e^x = k \implies x = \ln k$$ Since $k$ is 1 or 2, $\ln 1 = 0$ and $\ln 2 \approx 0.693$. 5. **Volume integral for each $k$:** - For $k=1$, bounds are $x=0$ to $x=\ln 3$. - For $k=2$, bounds are $x=\ln 2$ to $x=\ln 3$. 6. **Calculate volume for $k=1$:** $$V = \pi \int_0^{\ln 3} (e^x - 1)^2 \, dx = \pi \int_0^{\ln 3} (e^{2x} - 2e^x + 1) \, dx$$ 7. **Integrate term-by-term:** $$\int e^{2x} dx = \frac{e^{2x}}{2}, \quad \int e^x dx = e^x, \quad \int 1 dx = x$$ 8. **Evaluate the integral:** $$\int_0^{\ln 3} (e^{2x} - 2e^x + 1) dx = \left[ \frac{e^{2x}}{2} - 2e^x + x \right]_0^{\ln 3}$$ 9. **Calculate at bounds:** At $x=\ln 3$: $$\frac{e^{2\ln 3}}{2} - 2e^{\ln 3} + \ln 3 = \frac{(e^{\ln 3})^2}{2} - 2 \cdot 3 + \ln 3 = \frac{3^2}{2} - 6 + \ln 3 = \frac{9}{2} - 6 + \ln 3 = 4.5 - 6 + \ln 3 = -1.5 + \ln 3$$ At $x=0$: $$\frac{e^{0}}{2} - 2e^{0} + 0 = \frac{1}{2} - 2 + 0 = -1.5$$ 10. **Subtract:** $$(-1.5 + \ln 3) - (-1.5) = \ln 3$$ 11. **Multiply by $\pi$:** $$V = \pi \ln 3$$ 12. **Check for $k=2$ (just for confirmation):** Bounds: $x=\ln 2$ to $x=\ln 3$. Integral: $$V = \pi \int_{\ln 2}^{\ln 3} (e^x - 2)^2 dx = \pi \int_{\ln 2}^{\ln 3} (e^{2x} - 4e^x + 4) dx$$ Integrate: $$\left[ \frac{e^{2x}}{2} - 4e^x + 4x \right]_{\ln 2}^{\ln 3}$$ Evaluate at $x=\ln 3$: $$\frac{9}{2} - 12 + 4 \ln 3 = 4.5 - 12 + 4 \ln 3 = -7.5 + 4 \ln 3$$ At $x=\ln 2$: $$\frac{4}{2} - 8 + 4 \ln 2 = 2 - 8 + 4 \ln 2 = -6 + 4 \ln 2$$ Subtract: $$(-7.5 + 4 \ln 3) - (-6 + 4 \ln 2) = -7.5 + 4 \ln 3 + 6 - 4 \ln 2 = -1.5 + 4 (\ln 3 - \ln 2) = -1.5 + 4 \ln \frac{3}{2}$$ This is not equal to $\ln 3$, so $k=2$ is not the solution. **Final answer:** $k = 1$.