91Ó°ÊÓ

Sketch the region bounded by the function \(f(x)=x\ln x\) and the \(x\)-axis on the interval \([1, e^2]\). This will help you visualize the solid generated when the region is revolved around the \(x\)-axis.

Using the disk method, the volume of the solid is found by integrating the cross-sectional area of the solid along the axis of rotation. In this case, the cross-sectional area is given by \(A(x) = \pi (f(x))^2\) and the thickness of each disk is \(dx\). Therefore, the volume \(V\) can be written as $$V = \int_{1}^{e^2} A(x) dx = \int_{1}^{e^2} \pi (f(x))^2 dx.$$

Now we substitute the given function \(f(x) = x \ln x\) into the integral: $$V = \int_{1}^{e^2} \pi (x \ln x)^2 dx.$$

Expand the square of the function inside the integral: $$V = \int_{1}^{e^2} \pi x^2 (\ln x)^2 dx.$$

To find the volume, we need to evaluate the integral: $$V = \pi \int_{1}^{e^2} x^2 (\ln x)^2 dx.$$ To solve this integral, we will need to use integration by parts twice. Let \(u = (\ln x)^2\) and \(dv = x^2 dx\). Then, \(du = 2\ln x (\frac{1}{x}) dx = 2 \ln x \cdot x^{-1} dx\) and \(v = \frac{1}{3}x^3\). Now we can apply integration by parts formula: $$\int u dv= uv - \int v du.$$ Substitute the variables and get: $$V = \pi \left[\left. \frac{1}{3}x^3 (\ln x)^2 \right|_1^{e^2} - \int_{1}^{e^2} \frac{1}{3}x^3 (2 \ln x \cdot x^{-1}) dx \right].$$ Simplify the integral and get: $$V = \pi \left[\left. \frac{1}{3}x^3 (\ln x)^2 \right|_1^{e^2} - \frac{2}{3} \int_{1}^{e^2} x^2 \ln x dx\right].$$ Now, we apply integration by parts again. This time, let \(u = \ln x\) and \(dv = x^2 dx\). Then, \(du = (\frac{1}{x})dx\) and \(v = \frac{1}{3}x^3\). Apply the integration by parts formula again: $$V = \pi \left[\left. \frac{1}{3}x^3 (\ln x)^2 \right|_1^{e^2} - \frac{2}{3}\left(\left. \frac{1}{3}x^3 \ln x \right|_1^{e^2} - \int_{1}^{e^2} \frac{1}{3}x^3 (\frac{1}{x}) dx\right)\right].$$ Simplify the integral and get: $$V = \pi \left[\left. \frac{1}{3}x^3 (\ln x)^2 \right|_1^{e^2} - \frac{2}{3}\left(\left. \frac{1}{3}x^3 \ln x \right|_1^{e^2} - \frac{1}{3}\int_{1}^{e^2} x^2 dx\right)\right].$$

Evaluate the remaining integral and the expressions containing the limits: $$V = \pi \left[\frac{1}{3}(e^2)^3 (\ln (e^2))^2 - \frac{1}{3}(1)^3 (\ln (1))^2 - \frac{2}{3}\left(\frac{1}{3}(e^2)^3 \ln (e^2) - \frac{1}{3}(1)^3 \ln (1)\right) - \frac{2}{9}\int_{1}^{e^2} x^2 dx\right].$$ Evaluate the integral: $$V = \pi \left[ \frac{1}{3}(e^2)^3 (2)^2 - \frac{2}{3}\left(\frac{1}{3}(e^2)^3 (2)\right) - \frac{2}{9}\left(\frac{1}{3}(e^2)^3 - \frac{1}{3}(1)^3 \right)\right].$$ Simplify the expression: $$V = \pi \left[ \frac{4}{3}e^6 - \frac{4}{3}e^6 + \frac{2}{27} e^6 - \frac{2}{27} \right].$$

Combine the terms and write the final expression for the volume: $$V = \pi \left[ \frac{2}{27} e^6 - \frac{2}{27} \right].$$ This is the volume of the solid generated when the region bounded by \(f(x) = x \ln x\) and the \(x\)-axis on \(\left[1,e^{2}\right]\) is revolved about the \(x\)-axis.