Problem 44 - Calculus Early Transcendentals

Question

For the following regions \(R\), determine which is greater - the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or about the y-axis. \(R\) is bounded by \(y=x^{2}\) and \(y=\sqrt{8 x}\)

Step-by-Step Solution

Answer

4 = \pi \left[\frac{16}{2} - \frac{3}{5}(4)^{5/3}\right] = \frac{32\pi}{15}\) #Result# After evaluating the integral for the rotation around both the x and y axes, we have found the following volumes: - Volume when revolved around the x-axis: \(V_x = \frac{8\pi}{5}\) - Volume when revolved around the y-axis: \(V_y = \frac{32\pi}{15}\) To determine which volume is greater, we can compare the expressions: \(\frac{8\pi}{5} \approx 1.6\pi\) \(\frac{32\pi}{15} \approx 2.13\pi\) Since \(V_y > V_x\), the volume of the solid formed when region \(R\) is revolved around the y-axis is greater than the volume formed when revolved around the x-axis.

Step 1: Find Intersection points

To find the intersection points of \(y = x^2\) and \(y = \sqrt{8x}\), we need to solve the two functions simultaneously: \begin{align*} x^2 &= \sqrt{8x} \\ x^4 &= 8x \\ x^4 - 8x &= 0 \\ x(x^3 - 8) &= 0 \end{align*} The intersection points are \(x = 0\) and \(x = 2\). Therefore, the region \(R\) is bounded by \(0 \le x \le 2\).

Step 2: Set up volume integral for x-axis rotation

To determine the volume when region \(R\) is revolved around the x-axis, we will use the washer method for volumes of revolution, which states that the volume \(V\) is given by: \(V = \pi \int_{a}^{b} [\text{outer radius}^2(x) - \text{inner radius}^2(x)] \, dx\) Here, the outer radius is the distance from the x-axis to the curve \(y = \sqrt{8x}\): \(R_{outer}(x) = \sqrt{8x}\). The inner radius is the distance from the x-axis to the curve \(y = x^2\): \(R_{inner}(x) = x^2\). Thus, the volume integral for x-axis rotation becomes: \(V_x = \pi \int_{0}^{2} [(\sqrt{8x})^2 - (x^2)^2] \, dx\)

Step 3: Evaluate volume integral for x-axis rotation

Now, we evaluate the integral: \(V_x = \pi \int_{0}^{2} [8x - x^4] \, dx = \pi \left[\frac{8x^2}{2} - \frac{x^5}{5}\right]_0^2 = \pi \left[8 - \frac{32}{5}\right] = \frac{8\pi}{5}\)

Step 4: Set up volume integral for y-axis rotation

Similarly, we can set up the volume integral for the rotation around the y-axis. To set up the integral, we need to change the given equations which are in terms of x, into equations in terms of y. We will rotate around the y-axis, so we have: \(x = y^{1/2}\) \(x = (y/8)^{1/3}\) Now, we use the washer method to find the volume when region \(R\) is revolved around the \(y\)-axis. The outer radius is now the distance from the \(y\)-axis to the curve \(y = x^2\) curve after \(x = y^{1/2}\): \(R_{outer}(y) = y^{1/2}\) The inner radius is the distance from the y-axis to the curve \(y = \sqrt{8x}\) curve after \(x = (y/8)^{1/3}\): \(R_{inner}(y) = (y/8)^{1/3}\) Thus, the volume integral for y-axis rotation becomes: \(V_y = \pi \int_{0}^{4} [(y^{1/2})^2 - ((y/8)^{1/3})^2] \, dy\)

Step 5: Evaluate volume integral for y-axis rotation

Now, we evaluate the integral: $V_y = \pi \int_{0}^{4} [y - (y/8)^{2/3}] \, dy = \pi \left[\frac{y^2}{2} - \frac{3}{5}(y/8)^{5/3}\right]_0^-offsetof

Key Concepts

Volume of Revolution

When you revolve a region around a line, a 3D solid is formed. This is known as the volume of revolution. Imagine taking a 2D shape and spinning it around an axis like a potter shaping clay on a wheel. This process generates a symmetrical, three-dimensional object. In the case provided, the region between two curves is revolved around either the x-axis or the y-axis, resulting in two different volumes.

To find which volume is greater, one must calculate the solid’s volume for both axes of rotation. For finding these volumes, calculus methods such as the disk or washer method are employed. These methods rely on calculating the area of cross-sectional disks or washers, and then integrating along the axis of rotation. This approach not only simplifies the process but also provides a precise way to understand the structures involved in these 3D objects.

Washer Method

The washer method is a useful technique to determine the volume of a solid of revolution, especially when dealing with two curves. Imagine a washer, the flat ring-shaped object. When spinning a region bounded by two curves, a similar washer-like cross-section is formed. This approach takes into account the fact that the solid is hollow between two radii, an outer and an inner one.

• The **outer radius** is typically from the axis of rotation to the outer curve, in this scenario represented by the equation of the curve farther from the axis.
• The **inner radius** is from the axis of rotation to the inner curve.
  In the given problem, when rotating around the x-axis, the formula becomes: \(R_{outer}(x) = \sqrt{8x}\) and \(R_{inner}(x) = x^2\).

Inserting these into the washer formula for the x-axis revolution, you obtain: \[V = \pi \int [R_{outer}(x)^2 - R_{inner}(x)^2] dx\].
When applied correctly, this method efficiently captures the needed volume while accounting for the "hollow" nature of the solid.

Intersection Points

Discovering where two curves intersect can often determine the boundaries for integration when calculating such revolutions. Intersection points signify where the curves meet, and it's crucial to find them accurately, as they mark the limits of integration.

To solve for intersection points, equate the equations of the curves and solve for x. In this exercise, we have:

• Curve 1: \(y = x^2\)
• Curve 2: \(y = \sqrt{8x}\)

Set them equal: \(x^2 = \sqrt{8x}\), and solve to find the intersection points \(x = 0\) and \(x = 2\).
This means that the region of interest for integration is between these x-values. These points not only help in ensuring the right limits are chosen, but they also guide us in setting up the equations correctly for finding volumes or areas.

Integrals

Integrals are the heart of calculating volumes of revolution. They provide a way to sum up infinite small slices or washers into a comprehensive total volume.

In this context, once you have the intersection points and the radius expressions, the integral sums up those tiny volumes. Going through the specific exercise, you begin by setting up the integrals according to the washer method calculus formula:

• For x-axis: \[V_x = \pi \int_{0}^{2} [8x - x^4] \, dx\]
• For y-axis: \[V_y = \pi \int_{0}^{4} [y - (y/8)^{2/3}] \, dy\]

Evaluating these will give you the needed volume. The calculations often involve polynomial integration which, while straightforward, requires careful attention to detail. Completing these steps, students can confidently compare the resultant volumes from revolving around different axes.