Question

FInd the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(y=9-x^{2}(x \geq 0), x=0, y=0 ;\) about the line \(x=3\)

Short answer

The volume is \( 40.5\pi \).

Step-by-step solution

Sketch the Region R

The region R is bounded by the curves \( y = 9 - x^2 \), \( x = 0 \), and \( y = 0 \). Plotting these gives a region above the x-axis and to the right of the y-axis, under the parabola opening downwards with vertex at (0, 9) and intersection with x-axis at (3,0) and (-3,0) respectively. As \( x \geq 0 \), we only consider the part from (0,0) to (3,0).

Typical Rectangular Slice Representation

Imagine a vertical rectangular slice at a position \( x \) from the y-axis within the region R. The height of the slice is \( 9 - x^2 \), and its thickness is \( \Delta x \).

Shell Method Setup

When revolving around the line \( x = 3 \), the radius of the shell is \( 3 - x \). The approximate volume of a cylindrical shell is given by the formula \( 2\pi (\text{radius})(\text{height})(\text{thickness}) \). Thus, for our slice, the approximate volume becomes \( 2\pi (3-x)(9-x^2)\Delta x \).

Set Up the Integral

Integrate the approximate volume to find the total volume, from the lower bound \( x = 0 \) to the upper bound \( x = 3 \):\[V = \int_0^3 2\pi (3-x)(9-x^2)\, dx\]Simplifying, the integrand is \( 2\pi (27 - 9x^2 - 3x + x^3) \).

Evaluate the Integral

Evaluate the definite integral:\[V = 2\pi \int_0^3 (27 - 3x - 9x^2 + x^3) \, dx\]Integrating term by term:\[V = 2\pi \left[ 27x - \frac{3x^2}{2} - 3x^3 + \frac{x^4}{4} \right]_0^3\]Applying the bounds:\[V = 2\pi \left[ (81 - 27\cdot\frac{9}{2} - 3\cdot27 + 81/4) - 0 \right]\]Calculate:\[V = 2\pi \left( 81 - 40.5 - 81 + 20.25 \right)\]\[V = 2\pi \times 20.25\]\[V = 40.5\pi\]

Conclusion

The volume of the solid obtained by revolving the region around the line \( x=3 \) is \( 40.5\pi \).

Key concepts

Cylindrical Shell Method

The Cylindrical Shell Method is a technique used to find the volume of a solid of revolution, which is particularly useful when the disk method becomes cumbersome. This method involves revolving a region around an axis and imagining the solid as a series of coaxial cylindrical shells.

• Reason for Use: This method is advantageous when the axis of rotation is parallel to the axis of integration, like in our case, where the region bounded by the parabola is revolved around a vertical line, specifically, the line \(x = 3\).
• Shell Elements: Each cylindrical shell has a certain radius, height, and thickness. For our exercise, the radius of each shell is \(3 - x\), the height is \(9 - x^2\), and the thickness is an infinitesimal \(\Delta x\).
• Volume of a Shell: The formula for the volume of a shell is \(2\pi(\text{radius})(\text{height})(\text{thickness})\). This represents the surface area of the circular side of the cylindrical shell extended by its height.

Understanding the setup for each shell allows you to establish the integral that represents the entire volume of the solid.

Definite Integral

The definite integral is central to the process of finding the volume of solids generated by revolution. It sums up an infinite number of infinitesimally small quantities — in this case, the volumes of the cylindrical shells.

• Expression Setup: Starting from the approximate volume of a single shell, we create an integral that sums these volumes from the lower bound of \(x=0\) to the upper bound of \(x=3\). The integral is expressed as \[ V = \int_0^3 2\pi (3-x)(9-x^2) \, dx \].
• Integration Process: By expanding the expression \((3-x)(9-x^2)\), you simplify the problem into manageable polynomial terms. The definite integral takes this polynomial and computes the sum of the volumes of all the cylindrical shells.
• Importance of Bounds: The bounds of the integral, here from \(x=0\) to \(x=3\), define the limits between which the region R is considered. They are critical in determining the section of the curve you're working with.

The definite integral is powerful in solving practical problems involving areas, volumes, and other quantities that arise from continuous distributions.

Parabola

In this exercise, a parabola is one of the boundaries of the region being revolved. It's a simple yet powerful curve with unique properties.

• Description: The parabola is defined by the equation \(y = 9 - x^2\), indicating a downward-opening shape with the vertex at point \((0,9)\).
• Intersection Points: Since we consider \(x \geq 0\), the parabola intersects the x-axis at \((3,0)\). This intersection is vital because it sets the boundary for the region R within the first quadrant, from \(x=0\) to \(x=3\).
• Role in Volume: The height of the shell — a critical component in the shell method — is determined by the distance from the x-axis to the parabola, which is expressed as \(9 - x^2\).

Parabolas frequently appear in problems involving solids of revolution, and understanding their characteristics aids in setting up integrals accurately.

Axis of Revolution

The axis of revolution determines the direction and the way in which a region is revolved to create a solid. In this exercise, the axis is located at \(x=3\), which is crucial when applying the cylindrical shell method.

• Choice of Axis: The axis at \(x = 3\) means the region is being revolved around a vertical line that isn't the y-axis and is positioned to the right of the region's boundaries.
• Effect on Radius: This choice directly affects the radius of each cylindrical shell, calculated as the distance from the shell to the line of revolution. In our problem, this is \(3 - x\).
• Symmetry and Volume: The symmetry about the x-axis and the positioning of the axis of revolution ensure that the shells are generated consistently, simplifying the determination of volume through integration.

Choosing the correct axis of revolution allows for the accurate computation of volumes in complex geometries.