Problem 31 - Calculus and its applications

Question

Let \(R\) be the area bounded by the graph of \(y=9-x^{2}\) and the \(x\) -axis over [0,3] . a) Find the volume of the solid of revolution generated by rotating \(R\) around the \(x\) -axis. b) Find the volume of the solid of revolution generated by rotating \(R\) around the \(y\) -axis. c) Explain why the solids in parts (a) and (b) do not have the same volume.

Step-by-Step Solution

Answer

a) Volume around x-axis: \(129.6\pi\). b) Volume around y-axis: \(40.5\pi\). c) Different rotation axes create different volume elements.

Step 1: Understanding the Problem

The function given is \( y = 9 - x^2 \). We have to find the volumes of solids of revolution by rotating the area bounded by this function and the \( x \)-axis over the interval \([0, 3]\), first around the \( x \)-axis and then around the \( y \)-axis.

Step 2: Identify the Volume Formula for Rotation around X-axis

To find the volume of the solid when rotated around the \( x \)-axis, we use the disk method formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]where \( f(x) = 9 - x^2 \), \( a = 0 \), and \( b = 3 \).

Step 3: Calculate the Volume around the X-axis

Substitute \( f(x) = 9 - x^2 \) into the formula:\[ V = \pi \int_{0}^{3} (9 - x^2)^2 \, dx \]Evaluate the integral:\[ (9 - x^2)^2 = 81 - 18x^2 + x^4 \]So,\[ V = \pi \left[ \int_{0}^{3} 81 \, dx - 18 \int_{0}^{3} x^2 \, dx + \int_{0}^{3} x^4 \, dx \right] \]Calculating each integral:\[ \int_{0}^{3} 81 \, dx = 81x \bigg|_{0}^{3} = 243 \]\[ \int_{0}^{3} x^2 \, dx = \frac{x^3}{3} \bigg|_{0}^{3} = 9 \]\[ \int_{0}^{3} x^4 \, dx = \frac{x^5}{5} \bigg|_{0}^{3} = \frac{243}{5} \]Now substitute these back:\[ V = \pi (243 - 18 \times 9 + \frac{243}{5}) = \pi (243 - 162 + 48.6) = \pi (129.6) \]Thus, the volume is \( 129.6\pi \).

Step 4: Identify the Volume Formula for Rotation around Y-axis

To find the volume of the solid when rotating around the \( y \)-axis, use the shell method: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]where \( f(x) = 9 - x^2 \), \( a = 0 \), and \( b = 3 \).

Step 5: Calculate the Volume around the Y-axis

Substitute \( f(x) = 9 - x^2 \) into the shell formula:\[ V = 2\pi \int_{0}^{3} x(9 - x^2) \, dx = 2\pi \left[ \int_{0}^{3} 9x \, dx - \int_{0}^{3} x^3 \, dx \right] \]Evaluate each integral:\[ \int_{0}^{3} 9x \, dx = \frac{9x^2}{2} \bigg|_{0}^{3} = \frac{81}{2} \]\[ \int_{0}^{3} x^3 \, dx = \frac{x^4}{4} \bigg|_{0}^{3} = 20.25 \]Substitute these back:\[ V = 2\pi (40.5 - 20.25) = 2\pi (20.25) = 40.5\pi \]Thus, the volume is \( 40.5\pi \).

Step 6: Explain the Difference in Volumes

The volumes differ because rotating around the \( x \)-axis creates disks whose radius changes based on \( y = 9 - x^2 \), impacting the size more significantly, while rotating around the \( y \)-axis creates cylindrical shells whose radii \( x \) are smaller, less dominantly affecting the size of each infinitesimal element.

Key Concepts

Disk method

When calculating the volume of a solid formed by revolving a region about the x-axis, the disk method is a straightforward approach. The method visualizes the solid as a series of flat disks stacked along the axis of revolution. Each small segment of the solid can be thought of like a disk with a tiny thickness, \(dx\), and a radius equal to the value of the function at that point, \(f(x)\). This makes the calculation fairly intuitive when dealing with revolutions around the x-axis. The volume of each disk is \(\pi [f(x)]^2 dx\)

The formula used here is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

In the specific problem, the function is \(y = 9 - x^2\), and after setting up the integral, you solve to find the volume of the solid of revolution. The calculation includes expanding the square of the function and integrating term by term.
It's worth noting that the final volume depends on the boundaries, from \(a = 0\) to \(b = 3\). This boundary specifies the part of the curve being revolved, and the x-axis revolution simplifies using straightforward circular geometry, which makes the disk method here convenient.

Shell method

The shell method is another powerful technique for finding volumes of solids of revolution, especially useful when rotating a region about the y-axis. Unlike the disk method, which uses flat disks, the shell method involves cylinders with infinitesimal thickness stacked perpendicular to the axis of rotation. You visualize the solid as cylindrical shells with height equal to \(f(x)\) and radius equal to the x-coordinate.

The volume of each shell is calculated by considering the lateral surface area of the shell, which extends vertically from the axis of rotation and revolves around it. The formula for the shell method is:

• \( V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \)

In the context of the problem provided, revolving about the y-axis means using the shell method makes evaluating the integral simpler for the given function \(f(x) = 9 - x^2\). The integral ranges from \(a = 0\) to \(b = 3\) and integrates the product of \(x\) (the radius) and \(f(x)\) (the height).
The shell method is markedly different from the disk method, particularly in how it handles the dimensions affected by rotation. This method is advantageous when the region is bounded in a way that makes visualization and integral setup around a vertical axis overly complex for disks.

Volumes of revolution

Volumes of revolution is a fundamental concept in calculus that involves creating a 3D shape by rotating a 2D region around an axis. Understanding the different methods like the disk and shell methods allows you to tackle various problems of this type with confidence and efficiency.

• The disk method is often preferred for rotations around horizontal lines, like the x-axis, where it simplifies the geometry into circular slices.
• The shell method is advantageous for rotations around vertical lines, like the y-axis, offering a seamless way to capture the radial and height dimensions.

In terms of practical application, each method provides an analytical approach to deriving the volume of complex shapes that are otherwise challenging to compute directly. Each method is chosen based on how straightforward the integral setup and solution will be based on the specific axis of rotation.

The difference in resulting volumes, such as those observed in rotating a region about different axes, highlights the significance of the axis on the shape and dimensions of the created solid. Angle and direction of rotation can drastically affect the solid's total volume, due to the fundamental geometry involved with different axes.