Problem 36 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=0, y=\ln x, y=2, x=0$$

Step-by-Step Solution

Answer

Based on the provided step by step solution, write a short answer for the given exercise: To find the volume of the solid generated when region R, which is bounded by the curves y=0, y=lnx, y=2, and x=0, is revolved about the y-axis, we use the disk method. By expressing the radius in terms of y (x = e^y), we set up the area of each disk (A = πe^(2y)) and integrate over the interval [0, 2]. The volume integral evaluates to V = (1/2)π[e^4 - 1].

Step 1: Sketch the region R and identify the boundaries

To begin, sketch a graph to understand the region being revolved about the y-axis. Plot the functions y=0, y=lnx, y=2, and x=0 on a coordinate grid. Observe that region R is enclosed by these four curves.

Step 2: Set up the disk method integral

For the disk method, we want to integrate the area of each cross-sectional disk formed by the rotation. The thickness of each disk is dy as we are revolving about the y-axis. So, we need to express the radius of the disks in terms of y. First, solve for x by inverting the function y=lnx: $$x=e^y$$ Now, consider that the thickness of the disk is dy (y-axis), so the radius is in terms of y. The area of each disk is given by: $$A = \pi r^2$$ The radius here is x, so the area will be: $$A = \pi (e^y)^2 = \pi e^{2y}$$

Step 3: Set up the limits of integration for the volume integral

We need to determine the limits of integration for our volume integral. We are given the boundaries y=0 and y=2; these will be our limits of integration.

Step 4: Calculate the volume integral using the disk method

Integrate the disk area A with respect to y over the interval [0, 2]: $$V = \int_0^2 \pi e^{2y} dy$$ To solve this integral, apply the substitution: Let $$u = 2y$$ ; then $$du = 2 dy$$ and $$dy = \frac{1}{2} du$$ Now replace y with u: $$V = \int_0^4 \pi e^u \frac{1}{2} du$$ $$V = \frac{1}{2} \pi \int_0^4 e^u du$$ Integrate e^u with respect to u: $$V = \frac{1}{2} \pi [e^u] |_0^4$$ Now, evaluate the integral with the limits: $$V = \frac{1}{2} \pi [e^4 - e^0]$$ $$V = \frac{1}{2} \pi [e^4 - 1]$$ This is the final expression for the volume of the solid generated when region R is revolved about the y-axis.