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

Answer: The volume of the solid is greater when region R is revolved about the y-axis.

Step 1: Find the intersection points of the given functions

Set the given functions equal to each other and solve for x: \(x^2 = \sqrt{8x}\). Square both sides: \(x^4 = 8x\). Rearrange the equation to find x: \(x^4 - 8x = 0\). Factor out x: \(x(x^3 - 8) = 0\). Now, we get two intersection points \(x = 0\) and \(x = 2\). These will be the limits of integration for the next steps.

Step 2: Find the volume of the solid when rotated around the x-axis

The washer method uses the formula: \(V_x = \pi\int_a^b (R^2 - r^2) dx\) In this case, when revolving around the x-axis, \(R\) is the distance from the x-axis to the curve \(y=\sqrt{8x}\), and \(r\) is the distance from the x-axis to the curve \(y=x^2\). We first need to solve for \(R\) and \(r\) in terms of x: \(R=\sqrt{8x}\) and \(r=x^2\). Now, we can plug in the limits of integration found in step 1 and the values of \(R\) and \(r\) into the formula: \(V_x = \pi\int_0^2 (\sqrt{8x})^2 - (x^2)^2 dx\) Simplify and integrate: \(V_x = \pi\int_0^2 (8x - x^4) dx\) \(V_x = \pi[\frac{4x^2}{2} - \frac{x^5}{5}]_0^2\) \(V_x = \pi[\frac{4(2)^2}{2} - \frac{(2)^5}{5} - 0]\) \(V_x = \pi[\frac{32}{5}]\)

Step 3: Find the volume of the solid when rotated around the y-axis

Similarly as in step 2, we use the washer method, except this time we'll integrate with respect to y. We need to rewrite the expressions as functions of y: \(x = \frac{y}{8}\) and \(x = \sqrt{y}\). Now, we can find the R and r values in terms of y: \(R = \sqrt{y}\) and \(r = \frac{y}{8}\). We will also need the limits of integration in terms of y, which we can find by plugging our intersection points into one of the original equations: \(y_1 = (0)^2 = 0\) \(y_2 = (2)^2 = 4\) Now we can use the washer method to find the volume when rotated around the y-axis: \(V_y = \pi\int_0^4 (\sqrt{y})^2 - (\frac{y}{8})^2 dy\) \(V_y = \pi\int_0^4 (y - \frac{y^2}{64}) dy\) Integrate: \(V_y = \pi[\frac{y^2}{2} - \frac{y^3}{192}]_0^4\) \(V_y = \pi[\frac{(4)^2}{2} - \frac{(4)^3}{192} - 0]\) \(V_y = \pi[\frac{32}{3}]\)

Step 4: Compare the volumes

Now that we have the volumes for both the x-axis and y-axis rotations, we can compare them to see which is greater: \(V_x = \pi[\frac{32}{5}]\) \(V_y = \pi[\frac{32}{3}]\) Since \(\frac{32}{3} > \frac{32}{5}\), we can conclude that the volume of the solid generated when region R is revolved about the y-axis is greater than the volume of the solid generated when it is revolved about the x-axis.

Key Concepts

Washer Method

The washer method is a powerful tool used to calculate the volume of solids when a region is revolved around an axis. It involves finding the area of washers, or rings, formed by two functions. These washers have an outer radius and an inner radius, represented by the functions' values.
The volume is determined by integrating the areas of these washers over an interval. This requires knowledge of:

• Outer radius ((R))
• Inner radius ((r))

In our example, when revolving region \(R\) around the x-axis, the outer radius is \(\sqrt{8x}\) and the inner radius is \(x^2\). For the y-axis, these roles are reversed with expressions updated in terms of \(y\).
This method efficiently handles volumes where holes appear in the solid due to the intersection of curves.

Intersection Points

Finding intersection points is vital for setting the integration limits in volume calculations. It involves finding where the curves meet, which serves as boundaries for the region of interest.
By setting functions equal, we solve for \(x\) or \(y\), depending on the axis of rotation.

• In our problem, solve \(x^2 = \sqrt{8x}\) to get \(x = 0\) and \(x = 2\).

These values are crucial as they indicate where the solid begins and ends as it revolves around the axis, ensuring proper volume calculation.

Integration Limits

Integration limits define the boundaries over which you integrate to find the volume. These values come from the intersection points of the functions.
When revolving around the x-axis, you use those intersection \(x\)-values directly for the limits.

• For the y-axis, convert them into corresponding \(y\)-values.
• In this exercise, we found \(x = 0\) and \(x = 2\) which set the limits for \(x\)-axis integration from 0 to 2, and by converting these for \(y\), limits from 0 to 4 for the y-axis.

Correct limits ensure the integration captures the entire region and precisely calculates the volume of the solid.

Revolution About the x-Axis

Revolution about the x-axis transforms the region into a three-dimensional object. The resulting solid's volume depends on revolving curves around this axis.

• The formula used is \( V_x = \pi \int_a^b (R^2 - r^2) \, dx \).
• This represents the integral of the difference between the squares of outer and inner radii.

Observe how each curve's distance from the x-axis plays into calculating these radii.
Calculating the volume this way helps visualize how an area grows into a 3D shape, especially when grappling with abstract math problems.

Revolution About the y-Axis

Revolution about the y-axis involves rotating the region to create a solid around this vertical line. It requires rearranging functions so that \(x\) is expressed in terms of \(y\).

• For our problem, update equations to \(x = \frac{y}{8}\) and \(x = \sqrt{y}\).
• The integration formula becomes \( V_y = \pi \int_c^d (R^2 - r^2) \, dy \).

In this setup, limits change to \(y\) values, altering the way integration is performed.
This approach reveals how different axes of rotation affect the resulting solid's volume, offering a deeper understanding of geometric transformations.