Calculate the volumes of solids of rotation with upper and lower bounds with the shell method calculator. Know the whole computation procedure.
To use the shell method calculator, you will have to:
If you don’t have any questions, use the examples given below. Or you can use them to practice more about this method.
The shell method is used in calculus to determine the volume of a solid of revolution. When a region in the plane is revolved around a line, this method views the volume as being composed of a series of cylindrical shells. For a function y = f(x) bounded by x = a and x = b, and revolved around the y-axis, volume V is given by:
V = 2π a∫b xf(x) dx
In this formula:
The integral then sums the volumes of these shells over the interval [a, b].
It's worth noting that the shell method can also be applied when revolving around other axes, such as the x-axis, and the formula would be adjusted accordingly.
The shell method is utilized in calculus primarily to find the volume of solids of revolution. A solid of revolution is created by revolving a region in the plane about a line, which becomes the axis of rotation.
The shell method is especially useful when this region is bounded by vertical lines, as the method relies on visualizing the volume as a series of cylindrical shells formed by these vertical slices.
For instance, consider a curve or function defined on a certain interval, and you wish to revolve it around a vertical line, such as the y-axis. The shell method would be an intuitive approach for this, as each vertical slice (or "shell") of the region gets revolved around the axis to form a cylinder.
By summing up the volumes of all these infinitesimally thin cylinders—achieved through integration—you determine the total volume of the solid.
Citation (By Blacklemon67 at English Wikipedia, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=51359225)
In addition, the shell method proves beneficial when the solid has a hole in it or is hollow, or when the region being revolved isn't directly adjacent to the axis of rotation.
The method allows for a more straightforward setup in these situations than alternative techniques like the disk/washer method.
Identify the Region: Determine the area bounded by the curves you're revolving around an axis.
Visualize the Cylindrical Shell: Take a thin vertical strip (rectangle) of width
Δx from the bounded region. When this strip is revolved around the axis, it forms a cylindrical shell.
Determine Shell's Dimensions: The height of the shell is given by the function value f(x), and its average radius is x (the distance from the y-axis). The thickness of this shell is Δx, which will become dx in our integral.
Set Up the Integral: The volume of one shell is approximately 2πxf(x)Δx. To find the total volume, integrate this expression over the given interval.
Find the volume of the solid formed when the region bounded by y = √x and y = x (from x = 0 to x = 1) is revolved around the y-axis.
Solution:
Identify the Region: The area between the curves y = √x and y = x from x = 0 to x = 1 is our region of interest. Since y = √x lies above y = x in this interval, our representative rectangle's height will be √x − x.
Visualize the Shell: A vertical strip in this region, when revolved around the y-axis, will create our shell.
Determine the Dimensions:
Height of the shell: √x − x
Average radius: x
Thickness: dx
Set Up the Integral: The volume V is: V = 2π∫ x(√x− x) dx
Now, let's evaluate the integral:
V = 2π 0∫1 (x3/2 − x2) dx
V = 2π[ 2x5/2/5 - x3/3 ]01
V = 2π[ 2/5 - 1/3 ]
V = 2π( 6−5 / 15 )
V= 2π / 15
Determine the volume of the solid obtained by revolving the region bounded by the curve y = x2, the line x = 1, the x-axis, and the y-axis around the y-axis using the shell method. This region is a parabolic section from the origin (0,0) to the point (1,1).
Solution:
Identify the Region: The curve y = x2 from x = 0 to x = 1 forms our region of interest.
Visualize the Shell: Take a thin vertical strip of width Δx from the curve y = x2. This strip, when revolved around the y-axis, creates a cylindrical shell.
Determine Shell's Dimensions:
Height of the shell: x2
Average radius: x
Thickness: dx
Set Up the Integral:
The volume V of one shell is approximately 2πxx2 Δx. To find the total volume, integrate this over the interval from 0 to 1:
V = 2π 0∫1 x3 dx
Now, solve the integral:
V = 2π[ x4 / 4 ]01
V = 2π[ ¼ - 0]
V = 2π
While the method itself is abstract and primarily taught in academic settings, the principles behind it have applications in various real-life scenarios. Here are some daily life applications:
Manufacturing: In industries where hollow objects are produced, understanding the volume and thickness of the material is essential. This could be applicable in creating pipes, containers, bottles, or even certain auto parts.
Civil Engineering: When designing archways, tunnels, or cylindrical structures, engineers need to calculate the amount of material required. The shell method provides a way to determine the volume of such irregular shapes.
Medical Imaging: In techniques like MRI and CT scans, slices of the human body are taken and then combined to form a three-dimensional image. The principles of the shell method can be applied to understand the volume of certain organs or tumors based on these slices.
Food Industry: Think of a chocolate shell dessert or a hollow bread. Understanding the volume and thickness of the shell can help in maintaining consistency in production.
Oil and Gas Industry: Storage tanks often have cylindrical shapes. When these tanks have irregularities or are not perfect cylinders, the shell method can help determine the volume of fluid they can hold.
