# 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 ∫^{b}_{a} 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 ∫^{b}_{a} 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 ∫^{b}_{a} [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 ∫^{b}_{a} [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 ∫^{b}_{a} (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 ∫^{b}_{a} (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) = 3x^{3 }+ 7x^{5} – 12x^{2}
– 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) = 3x^{3 }+ 7x^{5} –
12x^{2} – 2x + 5

Integrating variable = x

2πx ∫^{b}_{a} f(x) dx = 2πx ∫^{2}_{1}
[3x^{3 }+ 7x^{5} – 12x^{2} – 2x + 5] dx

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

2πx ∫^{b}_{a} f(x) dx = 2π ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 5x] dx

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

2π ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 5x] dx = 2π [∫^{2}_{1} [3x^{4}]
dx^{ }+ ∫^{2}_{1} [7x^{6}] dx – ∫^{2}_{1}
[12x^{3}] dx – ∫^{2}_{1} [2x^{2}] dx + ∫^{2}_{1}
[5x] dx]

2π ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 5x] dx = 2π [3∫^{2}_{1} [x^{4}]
dx^{ }+ 7∫^{2}_{1} [x^{6}] dx – 12∫^{2}_{1}
[x^{3}] dx – 2∫^{2}_{1} [x^{2}] dx + 5∫^{2}_{1}
[x] dx]

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

2π ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 5x] dx = 2π [3 [x^{4+1} / 4 + 1]^{2}_{1}^{
}+ 7 [x^{6+1} / 6 + 1]^{2}_{1}^{ }– 12 [x^{3+1}
/ 3 + 1]^{2}_{1}^{ }– 2 [x^{2+1} / 2 + 1]^{2}_{1}^{
}+ 5 [x^{1+1} / 1 + 1]^{2}_{1}]

= 2π [3 [x^{5} / 5]^{2}_{1}^{
}+ 7 [x^{7} / 7]^{2}_{1}^{ }– 12 [x^{4}
/ 4]^{2}_{1}^{ }– 2 [x^{3} / 3]^{2}_{1}^{
}+ 5 [x^{2} / 2]^{2}_{1}]

= 2π [3/5 [x^{5}]^{2}_{1}^{
}+ 7/7 [x^{7}]^{2}_{1}^{ }– 12/4 [x^{4}]^{2}_{1}^{
}– 2/3 [x^{3}]^{2}_{1}^{ }+ 5/2 [x^{2}]^{2}_{1}]

= 2π [3/5 [x^{5}]^{2}_{1}^{
}+ [x^{7}]^{2}_{1}^{ }– 3 [x^{4}]^{2}_{1}^{
}– 2/3 [x^{3}]^{2}_{1}^{ }+ 5/2 [x^{2}]^{2}_{1}]

**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 [2^{5} – 1^{5}] +
[2^{7} – 1^{7}] – 3 [2^{4} – 1^{4}] – 2/3 [2^{3}
– 1^{3}] + 5/2 [2^{2} – 1^{2}]]

= 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 ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 5x] dx = 206.86 (3.14)

2π ∫^{2}_{1} [3x^{4
}+ 7x^{6} – 12x^{3} – 2x^{2} + 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) = 12y^{2 }+ 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) = 12y^{2 }+ 8y – 14

Integrating variable = y

2πy ∫^{b}_{a} f(y) dy = 2πy ∫^{3}_{1} [12y^{2
}+ 8y – 14] dy

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

2πy ∫^{b}_{a} f(y) dy = 2π ∫^{3}_{1}
[12y^{3 }+ 8y^{2} – 14y] dy

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

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [∫^{3}_{1} [12y^{3}]
dy^{ }+ ∫^{3}_{1} [8y^{2}] dy – ∫^{3}_{1}
[14y] dy]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [12∫^{3}_{1} [y^{3}]
dy^{ }+ 8∫^{3}_{1} [y^{2}] dy – 14∫^{3}_{1}
[14y] dy]

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

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [12 [y^{3+1} / 3 + 1]^{3}_{1}^{
}+ 8 [y^{2+1} / 2 + 1]^{3}_{1} – 14 [14y^{1+1}
/ 1 + 1]^{3}_{1}]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [12 [y^{4} / 4]^{3}_{1}^{
}+ 8 [y^{3} / 3]^{3}_{1} – 14 [y^{2} / 2]^{3}_{1}]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [12/4 [y^{4}]^{3}_{1}^{
}+ 8/3 [y^{3}]^{3}_{1} – 14/2 [y^{2}]^{3}_{1}]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [3 [y^{4}]^{3}_{1}^{
}+ 8/3 [y^{3}]^{3}_{1} – 7 [y^{2}]^{3}_{1}]

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

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [3 [3^{4} – 1^{4}] +
8/3 [3^{3} – 1^{3}] – 7 [3^{2} – 1^{2}]]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [3 [81 – 1] + 8/3 [27 – 1] – 7 [9 –
1]]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [3 [80] + 8/3 [26] – 7 [8]]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [240 + 208/3 – 56]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [240 + 69.33 – 56]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [309.33 – 56]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 2π [255.33]

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 510.67π

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

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 14y] dy = 510.67 (3.14)

2π ∫^{3}_{1} [12y^{3
}+ 8y^{2} – 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.