**Volumes**

## Volume of Solids by Slicing

Contents

So far in the Mathematics and Mathematics Extension 1 course, we have done volume of solids of revolution by rotating a curve about the or axes. We now extend this idea to include rotations about lines other than the coordinate axes. Students are required to derive and find the volume of solids from first principles.

When finding the volume of solids of revolution by the method of slicing, students need to understanding that the total volume of the solid is formed by summing up infinitely thin slices that are perpendicular to the axis of rotation.

### Students must show the following steps in order to get full marks

- Draw an accurate diagram
- Find an expression for the area of the cross-section of one slice
- Find an expression for the volume of a slice
- Find an expression for the total volume by summing up all the slices whilst taking the limit as the width approaches 0
- Solve the integral

The next example demonstrates what we already know from the Mathematics course.

**Example 1**

The region bounded by the curve , the -axis, and the lines and is rotated about the -axis. Find the volume of the solid of revolution formed.

### Solution 1

Area of cross section =

The following examples demonstrate how we can use the method of slicing to find the volume of the solid of revolution formed when a curve rotates about lines other than the coordinate axes.

**Example 2**

The region bounded by the curve and the -axis is rotated about the line . Find the volume of the solid of revolution.

### Solution 2

Area of cross section =

Area of cross section =

Note that upper bound is , because when ,

**Example 3**

The region bounded by the curve and the and axes is rotated about the line . Find the volume of the solid of revolution.

### Solution 3

Area of cross section =

Choose since we are using coordinate on the left branch

Area of cross section =

Note that upper bound is , because when ,

**Example 4**

[HSC 2010, Question 3]

The region bounded by the -axis and the curve is rotated about the line . Find the volume generated.

### Solution 4

Area of cross section =

and

Area of cross section =

Note that upper bound is , because when ,

**Example 5**

[HSC 2008, Question 5]

Let and be constants, with . A torus is formed by rotating the circle about the -axis.

The cross-section at , where , is an annulus. The annulus has inner radius and outer radius where and are the roots of

(i) Find and in terms of .

(ii) Find the area of the cross-section at height , in terms of .

(iii) Find the volume of the torus.

### Solution 5

i)

Note that

ii)

Area of cross section =

=

iii)

Note that the definite integral represents the area in a semicircle with radius b

**Example 6**

(i) lies on the hyperbola . Q is the foot of the perpendicular from P to the line . If and , show that .

(ii) The region bounded by the hyperbola and the line is rotated through about the line . Find the volume of the solid of revolution by slicing perpendicular to the axis of rotation.

### Solution 6

i)

Using the formula for distance from a line:

But

ii)

Area of cross-section

Note that varies from to (this correspond the to the intersection points of and , and and and their distance from the origin)

=

## Volume of Solids by Cylindrical Shells

Another method of finding the volume of a solid of revolution is by summing up cylindrical shells that are parallel to the axis of rotation. The reason behind multiple methods is because often one method gives elegant solutions to problems which would be challenging to solve by another.

There are two methods of using cylindrical shells:

- The first principle method where the volume of the solid of revolution is found by summing up thin cylindrical shells.
- Treating the cylindrical shell as a rectangular prism by rolling out the cylinder.

Both methods are acceptable by the syllabus but the second method tends to be easier, thus it will be used more often.

The following examples demonstrate how we can use the method of cylindrical shell to find the volume of the solid of revolution.

**Example 7**

[HSC 2007, Question 3]

Use the method of cylindrical shells to find the volume of the solid formed when the shaded region bounded by

is rotated about the -axis.

### Solution 7

Consider a thin cylindrical shell parallel to axis of rotation. Since the shell has very small width, its volume can be approximated as a rectangular prism. For any arbitrary point (x,y) on the curve, the resulting prism will have:

**Example 8**

[HSC 2009, Question 3]

The diagram shows the region enclosed by the curves and .

The region is rotated about the-axis.

Use the method of cylindrical shells to find the volume of the solid formed.

### Solution 8

(We will use first principles method here)

Consider a very thin cylindrical shell,

where is the coordinate of the parabola, and is the coordinate for the straight line.

Because is so small, and taking its square makes it even smaller, we approximate .

(

**Example 9**

[HSC 2006, Question 5]

A solid is formed by rotating the region bounded by the curve and the line about the -axis. Use the method of cylindrical shells to find the volume of this solid.

### Solution 9

**Example 10**

[HSC 2005, Question 4]

The shaded region between the curve , the -axis, and the lines and, where , is rotated about the -axis to form a solid of revolution.

(i) Use the method of cylindrical shells to find the volume of this solid in terms of .

(ii) What is the limiting value of this volume as ?

### Solution 10

i)

ii)

**Example 11**

[HSC 2011, Question 7]

The diagram shows the graph of for .

The area bounded by , the line and the -axis is rotated about the line to form a solid.

Use the method of cylindrical shells to find the volume of the solid.

### Solution 11

**Example 12**

The region bounded by the curve and the and -axes is rotated about the line . Find the volume of the solid of revolution.

### Solution 12

## Volume of Solids with Parallel Cross Sections of Similar Shapes

So far we have studied volumes by rotating a 2-dimensional region about a line. Now we will look at volumes of solids of irregular shape but with similar cross sections.

### Students must show the following steps in order to get full marks

- Draw an accurate diagram
- Find an expression for the area of an arbitrary cross-section
- Find an expression for the volume of the cross-section
- Find an expression for the total volume by summing up all the slices of the cross-section whilst taking the limit as the width approaches 0
- Solve the integral

The following examples demonstrate this concept.

**Example 13**

[HSC 2006, Question 4]

The base of a solid is the parabolic region shaded in the diagram.

Vertical cross-sections of the solid perpendicular to the -axis are squares.

Find the volume of the solid.

### Solution 13

Area of cross-section =

Volume of cross-section:

**Example 14**

[HSC 2011, Question 3]

The base of a solid is formed by the area bounded by and for .

Vertical cross-sections of the solid taken parallel to the -axis are in the shape of isosceles triangles with the equal sides of length 1 unit as shown in the diagram.

Find the volume of the solid.

### Solution 14

Area of cross-section =

Volume of cross-section:

**Example 15**

[HSC 2007, Question 4]

The base of a solid is the region bounded by the curve , the -axis and the lines and , as shown in the diagram.

Vertical cross-sections taken through this solid in a direction parallel to the -axis are squares. A typical cross-section, PQRS, is shown.

Find the volume of the solid.

### Solution 15

Area of cross-section =

Volume of cross-section:

**Example 16**

[HSC 2010, Question 6]

The diagram shows the frustum of a right square pyramid. (A frustum of a pyramid is a pyramid with its top cut off.)

The height of the frustum is m. Its base is a square of side m, and its top is a square of side m (with).

A horizontal cross-section of the frustum, taken at height m, is a square of side m, shown shaded in the diagram.

(i) Show that .

(ii) Find the volume of the frustum.

### Solution 16

i)

Consider the triangle formed from the edges, as shown below:

Noting that they are similar triangles, we have:

ii)

Area of cross-section =

Volume of cross-section:

**Example 17**

[HSC 2009, Question 6]

The base of a solid is the region enclosed by the parabola and the -axis. The top of the solid is formed by a plane inclined at to the -plane. Each vertical cross-section of the solid parallel to the -axis is a rectangle. A typical cross-section is shown shaded in the diagram.

Find the volume of the solid.

### Solution 17

Area of cross-section =

Volume of cross-section:

**Example 18**

[HSC 2005, Question 5]

The base of a right cylinder is the circle in the -plane with centre and radius 3. A wedge is obtained by cutting this cylinder with the plane through the -axis inclined at to the -plane, as shown in the diagram.

A rectangle slice ABCD is taken perpendicular to the base of the wedge at a distance from the -axis.

(i) Show that the area of ABCD is given by .

(ii) Find the volume of the wedge.

### Solution 18

i)

Area =

ii)

Volume of cross-section: