What is the shell method and how to calculate it?

 What is the shell method and how to calculate it?



In calculus, the shell method is a well-known method for calculating the volume of solids of revolutions. Rather than consider horizontal slices, it considers vertical slices, which simplifies certain problems where vertical slices are easier to describe.

It is dependent on the axis to make the shells. In this post, we are going to explain the shell method along with its definition, formulas, and examples.

What is the shell method?

In calculus, the process of determining the volumes by disintegrating a solid of revolution into cylindrical shells is said to be the shell method. It is totally dependent on the axis of the region according to the vertical strips.

Rather than rotating rings or disks, we will rotate cylinders in this method. It is simply an alternative to the disk method and washer method as these methods rotate the body in rings and disks while the shell method helps to rotate it in cylinders.

Formulas of shell method

There are different formulas for the shell method depending on the axis of rotation.

Rotation about y-axis

A solid's volume is obtained by rotating it around the y-axis by using the y = f(x) from a to b such as:

Volume = V = 2πx ba f(x) dx

Rotation about x-axis

A solid's volume is obtained by rotating it around the x-axis by using the x = f(y) from a to b such as:

Volume = V = 2πy ba f(y) dy

Rotation between two curves about the y-axis

A solid's volume is obtained by rotating it between two curves around the y-axis by using the f(x) and g(x) from a to b such as:

Volume = V = 2πx ba [f(x) – g(x)] dx

Rotation between two curves about the x-axis

A solid's volume is obtained by rotating it between two curves around the x-axis by using the f(y) and g(y) from a to b such as:

Volume = V = 2πy ba [f(y) – g(y)] dy

 

Rotation between two curves about x = h

A solid's volume is obtained by rotating it between two curves around the x = h by using the f(x) and g(x) from a to b such as:

Volume = V = 2π ba (x – h) [f(x) – g(x)] dx

Rotation between two curves about y = k

A solid's volume is obtained by rotating it between two curves around the x-axis by using the f(y) and g(y) from a to b such as:

Volume = V = 2π ba (y – k) [f(y) – g(y)] dy

 

How to calculate the shell method?

The volume of the solid of revolution can be calculated easily with the help of shell method formulas according to the rotation of curves. A shell method calculator is a helpful resource to calculate the shell method online to avoid time consuming calculations.

Here are a few examples to learn how to calculate the shell method manually.

Example 1: Rotation about the y-axis

Calculate the volume of the solid of revolution about the y-axis

f(x) = 3x3 + 7x5 – 12x2 – 2x + 5 in the interval of [1, 2]

Solution

Step 1: First of all, take the general formula of the shell method about the y-axis and write the function to the formula.

f(x) = 3x3 + 7x5 – 12x2 – 2x + 5

Integrating variable = x

2πx ∫ba f(x) dx = 2πx ∫21 [3x3 + 7x5 – 12x2 – 2x + 5] dx

Step 2: Now multiply the function f(x) by “x” which is outside the integral notation.

2πx ∫ba f(x) dx = 2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx

Step 3: Now apply the sum and difference rules of integration to write the integral notation with each function separately.

2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx = 2π [∫21 [3x4] dx + ∫21 [7x6] dx – ∫21 [12x3] dx – ∫21 [2x2] dx + ∫21 [5x] dx]

2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx = 2π [3∫21 [x4] dx + 7∫21 [x6] dx – 12∫21 [x3] dx – 2∫21 [x2] dx + 5∫21 [x] dx]

Step 4: Now integrate the above expression with the help of the power law of integration.

2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx = 2π [3 [x4+1 / 4 + 1]21 + 7 [x6+1 / 6 + 1]21 – 12 [x3+1 / 3 + 1]21 – 2 [x2+1 / 2 + 1]21 + 5 [x1+1 / 1 + 1]21]

= 2π [3 [x5 / 5]21 + 7 [x7 / 7]21 – 12 [x4 / 4]21 – 2 [x3 / 3]21 + 5 [x2 / 2]21]

= 2π [3/5 [x5]21 + 7/7 [x7]21 – 12/4 [x4]21 – 2/3 [x3]21 + 5/2 [x2]21]

= 2π [3/5 [x5]21 + [x7]21 – 3 [x4]21 – 2/3 [x3]21 + 5/2 [x2]21]

Step 5: Now apply the upper and lower limit to the above expression with the help of the fundamental theorem of calculus.

= 2π [3/5 [25 – 15] + [27 – 17] – 3 [24 – 14] – 2/3 [23 – 13] + 5/2 [22 – 12]]

= 2π [3/5 [32 – 1] + [128 – 1] – 3 [16 – 1] – 2/3 [8 – 1] + 5/2 [4 – 1]]

= 2π [3/5 [31] + [127] – 3 [15] – 2/3 [7] + 5/2 [3]]

= 2π [93/5 + 127 – 45 – 14/3 + 15/2]

= 2π [18.6 + 127 – 45 – 14/3 + 15/2]

= 2π [18.6 + 127 – 45 – 4.67 + 15/2]

= 2π [18.6 + 127 – 45 – 4.67 + 7.5]

= 2π [145.6 – 45 – 4.67 + 7.5]

= 2π [100.6 – 4.67 + 7.5]

= 2π [95.93 + 7.5]

= 2π [103.43]

= 206.86 π

Step 6: Now substitute the value of pi such as π = 3.14

2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx = 206.86 (3.14)

2π ∫21 [3x4 + 7x6 – 12x3 – 2x2 + 5x] dx = 649.54

Example 2: Rotation about the x-axis

Calculate the volume of the solid of revolution about the x-axis

f(y) = 12y2 + 8y – 14 in the interval of [1, 3]

Solution

Step 1: First of all, take the general formula of the shell method about the x-axis and write the function to the formula.

f(y) = 12y2 + 8y – 14

Integrating variable = y

2πy ∫ba f(y) dy = 2πy ∫31 [12y2 + 8y – 14] dy

Step 2: Now multiply the function f(y) by “y” which is outside the integral notation.

2πy ∫ba f(y) dy = 2π ∫31 [12y3 + 8y2 – 14y] dy

Step 3: Now apply the sum and difference rules of integration to write the integral notation with each function separately.

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [∫31 [12y3] dy + ∫31 [8y2] dy – ∫31 [14y] dy]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [12∫31 [y3] dy + 8∫31 [y2] dy – 14∫31 [14y] dy]

Step 4: Now integrate the above expression with the help of the power law of integration.

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [12 [y3+1 / 3 + 1]31 + 8 [y2+1 / 2 + 1]31 – 14 [14y1+1 / 1 + 1]31]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [12 [y4 / 4]31 + 8 [y3 / 3]31 – 14 [y2 / 2]31]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [12/4 [y4]31 + 8/3 [y3]31 – 14/2 [y2]31]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [3 [y4]31 + 8/3 [y3]31 – 7 [y2]31]

Step 5: Now apply the upper and lower limit to the above expression with the help of the fundamental theorem of calculus.

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [3 [34 – 14] + 8/3 [33 – 13] – 7 [32 – 12]]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [3 [81 – 1] + 8/3 [27 – 1] – 7 [9 – 1]]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [3 [80] + 8/3 [26] – 7 [8]]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [240 + 208/3 – 56]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [240 + 69.33 – 56]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [309.33 – 56]

2π ∫31 [12y3 + 8y2 – 14y] dy = 2π [255.33]

2π ∫31 [12y3 + 8y2 – 14y] dy = 510.67π

Step 6: Now substitute the value of pi such as π = 3.14

2π ∫31 [12y3 + 8y2 – 14y] dy = 510.67 (3.14)

2π ∫31 [12y3 + 8y2 – 14y] dy = 1603.49

Final words

Now you can grab all the basics of the shell method from this post. We have discussed the definition, formulas, and solved examples of the shell method to understand it more accurately. Now you are witnessed that this topic is not a difficult one.

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