# 4. Volume of Solid of Revolution by Integration

by M. Bourne

A lathe

Many solid objects, especially those made on a **lathe**, have a circular cross-section and curved sides. In this section, we see how to find the **volume** of such objects using integration.

Objects made on a lathe ...

### Example 1

### Applications

Don't miss the winecask and watermelon applications in this section.

Consider the area bounded by the straight line `y = 3x`, the `x`-axis, and `x = 1`:

When the shaded area is rotated 360° about the `x`-axis, a volume is generated.

The resulting solid is a cone:

Area under the curve `y=3x` from `x=0` to `x=1` rotated around the `x`-axis, showing a typical disk.

## Disk Method for Finding Volumes

To find this volume, we could take slices (the **dark green** disk shown above), each `dx` wide and radius `y`:

The **volume** of a cylinder is given by:

`V = πr^2h`

Because `"radius" = r = y` and each disk is `dx` high, we notice that the volume of each slice is:

`V = πy^2\ dx`

Adding the volumes of the disks (with infinitely small `dx`), we obtain the formula:

`V=pi int_a^b y^2dx` which means `V=pi int_a^b {f(x)}^2dx`

where:

`y = f(x)` is the equation of the curve whose area is being rotated`a` and `b` are the limits of the area being rotated

`dx` shows that the area is being rotated about the `x`-axis

**NOTE:** We use the **disk method** only (not **Shell Method**) in this section.

Applying the formula `V=pi int_a^b y^2dx` to the earlier example, we have:

`"Vol" =pi int_a^b y^2dx`

`=pi int_0^1(3x)^2dx`

`=pi int_0^1 9x^2dx`

`=pi[3x^3]_0^1`

`=pi[3]-pi[0]`

`=3pi\ "unit"^3`

CHECK: Does the method work? We can find the volume of the cone using

`"Vol"=(pi r^2h)/3`

` = (pi(3)^2(1))/3`

`=(9pi)/3`

`=3pi\ "unit"^3` (Checks OK.)

### Example 2

Find the volume if the area bounded by the curve `y = x^3+ 1`, the `x`-axis and the limits of `x = 0` and `x = 3` is rotated around the `x`-axis.

## Volume by Rotating the Area Enclosed Between 2 Curves

If we have 2 curves `y_2` and `y_1` that enclose some area and we rotate that area around the `x`-axis, then the volume of the solid formed is given by:

`"Volume"=pi int_a^b[(y_2)^2-(y_1)^2]dx`

In the following general graph, `y_2` is above `y_1`. The lower and upper limits for the region to be rotated are indicated by the vertical lines at `x = a` and `x = b`.

Area bounded by the curves `y_1` and `y_2`, & the lines `x=a` and `x=b`.

### Example 3

A cup is made by rotating the area between `y = 2x^2` and `y = x + 1` with `x ≥ 0` around the `x`-axis. Find the volume of the material needed to make the cup. Units are `"cm"`.

**Rotation around the ***y***-axis**

*y*

When the shaded area is rotated 360° about the `y`*-*axis, the volume that is generated can be found by:

`V=pi int_c^d x^2dy` which means `V=pi int_c^d {f(y)}^2dy`

where:

`x =f(y)` is the equation of the curveexpressed in terms of`y``c` and `d` are the upper and lower y limits of the area being rotated

`dy` shows that the area is being rotated about the `y`-axis

### Example 4

Find the volume of the solid of revolution generated by rotating the curve `y = x^3` between `y = 0` and `y = 4` about the `y`-axis.

## Exercises

Find the volume generated by the areas bounded by the given curves if they are revolved about the given axis:

**(1)** The straight line `y = x`, between `y = 0` and `x = 2`, revolved about the `x`-axis.

**(2)** The curve `y = 2x − x^2` bounded by `y = 0`, revolved about the `x`-axis.

**(3)** The curve `y^2 = x`, bounded by `y = 4` and `x = 0`, revolved about the `y`-axis.

**(4)** The curve `x^2 + 4y^2 = 4` in quadrant I, revolved around the `y`-axis.

## Applications

### 1. Volume of a wine cask

A wine cask has a radius at the top of 30 cm and a radius at the middle of 40 cm. The height of the cask is 1 m. What is the volume of the cask (in L), assuming that the shape of the sides is parabolic?

### 2. Volume of a watermelon

A watermelon has an ellipsoidal shape with major axis 28 cm and minor axis 25 cm. Find its volume.

**Historical Approach: **Before calculus, one way of approximating the volume would be to slice the watermelon (say in 2 cm thick slices) and add up the volumes of each slice using `V = πr^2h`.

Interestingly, Archimedes (the one who famously jumped out of his bath and ran down the street shouting "Eureka! I've got it") used this approach to find volumes of spheres around 200 BC. The technique was almost forgotten until the early 1700s when calculus was developed by Newton and Leibniz.

We see how to do the problem using both approaches.

Volume using historical method:

"Exact" Volume (using Integration):

[See also Archimedes and the area of the parabolic segment.]

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