Problem 45 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Let \(S\) be the solid generated when \(R\) is revolved about the given axis. If possible, find the volume of \(S\) by both the disk/washer and shell methods. Check that your results agree and state which method is easier to apply. \(y=\sqrt{\ln x}, y=\sqrt{\ln x^{2}},\) and \(y=1 ;\) revolved about the \(x\) -axis

Step-by-Step Solution

Answer

To determine the volume of the solid generated by revolving the region R, bounded by \(y=\sqrt{\ln x}, y=1\), and \(x=\sqrt{\ln x^2}\), around the x-axis, we can use either the disk/washer method or the shell method. We found that: 1. The region R is bounded by the curves from \(x=e\) to \(x=\sqrt{e}\). 2. Using the disk/washer method, the volume of the solid is \(\frac{\pi e}{2}\). 3. Using the shell method, the volume of the solid is also \(\frac{\pi e}{2}\). Both methods were successful in determining the volume. However, in this case, the disk/washer method was easier to apply, as it required fewer algebraic manipulations compared to the shell method.

Step 1: Find the limits of integration

To do this, let's first find where the first and third curves intersect: \(\sqrt{\ln x}=1\) Squaring both sides: \(\ln x = 1\) Now, by exponentiating both sides: \(x=e\) Next, let's find where the second and the third curves intersect: \(\sqrt{\ln x^2}=1\) Squaring both sides: \(\ln x^2 = 1\) Now by exponentiating both sides: \(x^2=e^1\Rightarrow x=\pm\sqrt{e}\) However, since we are looking at \(R\), \(x\ge0\), therefore \(x=\sqrt{e}\). So, the region R is bounded by the curves from \(x=e\) to \(x=\sqrt{e}\). **Step 2: Calculate the volume using the disk/washer method**

Step 2: Set up the disk/washer method integral

For this method, the volume of the solid generated by revolving the region R around the x-axis is given by the following formula: \(V=\pi\int_{a}^{b}[f_1^2(x)-f_2^2(x)]dx\) Where \(f_1(\-x) = \sqrt{\ln x}\) and \(f_2(x) = 1\), the outer and inner functions, respectively. The limits of integration, \(a=e\) and \(b=\sqrt{e}\). So, the integral is: \(V=\pi\int_{e}^{\sqrt{e}}[(\sqrt{\ln x})^2-1^2]dx\) **Step 3: Calculate the result of the integral**

Step 3: Compute the disk/washer method integral

Integrating, we have: $\begin{aligned} V&=\pi\int_{e}^{\sqrt{e}}(\ln x-1)dx \\ &= \pi\left[x\ln x-x\right]_{e}^{\sqrt{e}} \\ &= \pi\left[(\sqrt{e}\ln\sqrt{e}-\sqrt{e})-(e\ln e-e)\right] \\ &= \pi\left(\frac{e}{2}-e+1\right) \\ &= \frac{\pi e}{2} \end{aligned}$ So, using the disk/washer method, the volume of the solid is \(\frac{\pi e}{2}\). **Step 4: Calculate the volume using the shell method**

Step 4: Set up the shell method integral

For the shell method, the volume of the solid generated by revolving the region R around the x-axis is given as: \(V=2\pi\int_{c}^{d}r(y)h(y)dy\) Where \(r(y) = e^y\) and \(h(y) = \sqrt{e}-e^y\) are the radial and height functions, respectively, using the curves \(x=e^y\) and \(x=e^{2y}\). Also, \(c=\sqrt{\ln e}\) and \(d=\sqrt{\ln(\sqrt{e})}\). Thus, the integral is: \(V=2\pi\int_{1}^{0}e^{y}\left(\sqrt{e}-e^y\right)dy\) **Step 5: Calculate the result of the integral**

Step 5: Compute the shell method integral

Integrating the expression we have: $\begin{aligned} V&=2\pi\int_{1}^{0}e^{y}(\sqrt{e}-e^y)dy \\ &= 2\pi\left[-\sqrt{e}e^y+\frac{1}{2}e^{2y}\right]_{1}^{0} \\ &= 2\pi\left[-\sqrt{e}+\frac{1}{2}e-(1-\frac{1}{2})\right] \\ &= \pi\left(-2\sqrt{e}+e+1\right) \\ &= \frac{\pi e}{2} \end{aligned}$ So, using the shell method, the volume of the solid is also \(\frac{\pi e}{2}\). The results from both methods agree. While both methods yield the same result, the disk/washer method was easier to set up in this particular example because we could directly use the given functions for the outer and inner curves. The shell method required inversion of the functions and additional algebra steps, making it slightly more complicated to apply.

Key Concepts

Disk/Washer Method

Finding the volume of a solid of revolution can be made simpler using the Disk/Washer Method. This method involves slicing the solid perpendicular to the axis of rotation to create circular disks or washers.

The idea is to find the volume of each of these disks and sum them up from one end of the solid to the other. The formula for the volume of the disk is given by:

• \(V = \pi \int_{a}^{b} [f_{outer}^2(x) - f_{inner}^2(x)] \, dx\)

In this formula, \(f_{outer}(x)\) is the outer radius function, and \(f_{inner}(x)\) is the inner radius function.

When the inner radius is zero, it is called the Disk Method. If there is a hole, it becomes the Washer Method.

In our problem, we rotated about the x-axis, so the outer and inner radii are \(\sqrt{\ln{x}}\) and \(1\) respectively. By calculating the integral across the limits, we found the volume to be \(\frac{\pi e}{2}\). This method was more straightforward here as we had ready functions for the radii.

Shell Method

The Shell Method offers a different approach by summing up the cylindrical shells formed by slicing the solid parallel to the axis of rotation. Here, each slice is a "shell" that unwraps into a rectangle when rotated.

This method is particularly handy when the axis of rotation is vertical or when it's more cumbersome to use disks. The volume formula for the shell method is:

• \(V = 2\pi \int_{c}^{d} r(y) h(y) \, dy\)

Where:

• \(r(y)\) is the distance from the shell to the rotation axis
• \(h(y)\) is the height of the shell

In our exercise, for rotation about the x-axis, we transformed the curves for the shell formation. Calculating \(r(y)\) and \(h(y)\) involved using the curves \(x=e^y\) and \(x=e^{2y}\), resulting in a solution that also gave \(\frac{\pi e}{2}\). Since it required more algebraic manipulation than the Washer Method, the Shell Method wasn't as straightforward.

Limits of Integration

The Limits of Integration are critical for setting the range over which you evaluate the integrals when using both the disk/washer and shell methods for finding volumes.

These limits identify where the region of interest starts and ends along the axis being integrated. Typically, they are found by solving for the intersection points of the curves involved.

In the provided exercise, to determine these limits, we equate the given functions to the boundary condition \(y = 1\), leading to solving equations like \(\sqrt{\ln{x}} = 1\) and \(\sqrt{\ln{x^2}} = 1\). These computations resulted in \(x = e\) and \(x = \sqrt{e}\) as the bounds for the region of interest.

Thus, the limits \(a = e\) and \(b = \sqrt{e}\) are used for the disk/washer integral, while appropriate \(y\)-limits are set for the shell method, ensuring both methods cover the correct sections of the solid's boundary.