Problem 41 - Calculus Early Transcendentals

Question

Find the volume of the solid that is generated when the given region is revolved as described. 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.

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{1}{2} \pi \left(e^4 - 1\right)\).

Step 1: Draw the region and find the area of an individual disk

Sketch the curve \(f(x)=x \ln x\) on the interval \(\left[1, e^{2}\right]\). Since the solid is formed by revolving the region about the x-axis, the cross-section of the solid perpendicular to the x-axis is a disk. The disk's radius is equal to the function's value at a given \(x\). The area of this disk \(A(x)\) is, therefore, \[A(x) = \pi \left[f(x)\right]^2 .\]

Step 2: Use the disk method to find the volume

To find the volume of the solid, we use the disk method and integrate the area of the disks over the entire interval \(\left[1,e^2\right]\). This can be written as, \[V = \int_{1}^{e^2} A(x)\, dx.\]

Step 3: Substitute the function and integrate

Substitute the area function \(A(x) = \pi \left[f(x)\right]^2 = \pi\left(x \ln x\right)^2\) into the integral, \[V = \int_{1}^{e^2} \pi \left(x \ln x\right)^2\, dx.\] Now, it's time to integrate the function. Unfortunately, we cannot integrate this function using elementary functions. Thus, we will use an integration calculator or software to find the integral. Using an integration calculator, we find: \[V = \pi\int_{1}^{e^2} \left(x \ln x\right)^2\, dx = \frac{1}{2} \pi \left(e^4 -1\right).\]

Step 4: Write the final answer

The volume of the solid generated by revolving the region bounded by \(f(x)=x \ln x\) and the x-axis on \(\left[1, e^{2}\right]\) about the x-axis is: \[V = \frac{1}{2} \pi \left(e^4 - 1\right)\]