1. **Problem Statement:** We are given the graph of the function $y = 2 \ln x$ bounded by the line $y = 2$, the x-axis ($y=0$), and the y-axis ($x=0$). We need to sketch this graph and identify the representative strip for volume calculation when this area is rotated about the x-axis. 2. **Graph and Boundaries:** - The function is $y = 2 \ln x$. - The line $y=2$ intersects the curve where $2 \ln x = 2 \Rightarrow \ln x = 1 \Rightarrow x = e$. - The x-axis is $y=0$. - The y-axis is $x=0$, but note $\ln x$ is undefined at $x=0$, so the domain is $x>0$. 3. **Representative Strip:** - To find the volume generated by rotating the area about the x-axis, we consider a vertical strip at position $x$ with thickness $dx$. - The height of the strip is from $y=0$ to $y=2 \ln x$. - When rotated about the x-axis, this strip forms a disk with radius $y = 2 \ln x$. 4. **Volume Calculation Formula:** - The volume of the solid of revolution generated by rotating about the x-axis is given by the disk method: $$V = \pi \int_a^b [f(x)]^2 dx$$ - Here, $f(x) = 2 \ln x$, and the limits are from $x=1$ (where $y=0$) to $x=e$ (where $y=2$). 5. **Calculate the Volume:** $$V = \pi \int_1^e (2 \ln x)^2 dx = 4\pi \int_1^e (\ln x)^2 dx$$ 6. **Evaluate the integral $\int (\ln x)^2 dx$:** Use integration by parts: Let $u = (\ln x)^2$, $dv = dx$. Then $du = 2 \frac{\ln x}{x} dx$, $v = x$. $$\int (\ln x)^2 dx = x (\ln x)^2 - 2 \int \ln x dx$$ We know: $$\int \ln x dx = x \ln x - x + C$$ So, $$\int (\ln x)^2 dx = x (\ln x)^2 - 2 (x \ln x - x) + C = x (\ln x)^2 - 2x \ln x + 2x + C$$ 7. **Apply limits from 1 to $e$:** At $x=e$: $$e (\ln e)^2 - 2e \ln e + 2e = e (1)^2 - 2e (1) + 2e = e - 2e + 2e = e$$ At $x=1$: $$1 (\ln 1)^2 - 2(1) \ln 1 + 2(1) = 0 - 0 + 2 = 2$$ 8. **Calculate definite integral:** $$\int_1^e (\ln x)^2 dx = e - 2 = e - 2$$ 9. **Calculate volume:** $$V = 4 \pi (e - 2)$$ **Final answer:** $$\boxed{V = 4 \pi (e - 2)}$$