### Integration - Volume of Revolution

Suppose that f(x) is a function of x and the curve of the function is rotated around the x-axis through 360^{0}. It clearly forms
a solid object. The volume of the object can be calculated as follows:

Consider a small disk with thickness *dx*. If its volume is *dv,*

dv =π y^{2} dx

dv =π f(x)^{2} dx

In order to find the total volume, *V* of the object, let's integrate the above between *a* and *b*.

**V =π _{a} ∫ ^{b} f(x)^{2} dx.**

So, in order to find the **volume of revolution:**

1) Square the function and then integrate it

2) Then, multiply the answer byπ

**E.g.1**

Show that the volume of a cylinder is, **V** =π r^{2} h.

A cylinder is produced when the line **y = r** is rotated around x-axis.

V = π _{0}∫^{h} f(x)^{2} dx

V = π _{0}∫^{h} r^{2} dx

V = π [r^{2}x]^{h}_{0}

**V =π r ^{2} h.**

In the following applet, y = 2 line is rotated around the x-axis, shown in red. The y-axis is shown in green. Please press the *play* button or move the slider. You will see the cylinder being formed.

**E.g.2**

Show that the volume of a cone is, **V** = 1/3π r^{2} h.

A cone is produced when a line **y = mx** is rotated around the x-axis.

V =π _{0}∫^{h} f(x)^{2} dx

V = π _{0}∫^{h} (mx)^{2} dx

V =π m^{2} _{0} ∫ ^{h} x^{2} dx

V =π m^{2} [ x^{3} / 3]_{0}^{h}

m, the gradient of the line, = r /h

V =π r^{2} / h^{2} h^{3} / 3

**V = 1/3π r ^{2} h**

In the following applet, y = x line is rotated around the x-axis, shown in red. The y-axis is shown in green. Please press the *play* button or move the slider. You will see the cone being formed.

**E.g.3**

Show that the volume of a sphere is, **V** = 4/3π r^{3}.

A sphere is produced when the semi-circle, **x ^{2} + y^{2} = r^{2}** is rotated around the x-axis.

V =π _{-r}∫^{r} f(x)^{2} dx

V = π _{-r}∫^{r} [ √r^{2} - x^{2} ]^{2} dx

V = π _{-r}∫^{r}[r^{2} - x^{2}] dx

V =π [ r^{2}x - x^{3}/3 ]^{r}_{-r}

V =π [ r^{3} - r^{3}/3] - [-r^{3} - (-r)^{3}/3 ]

V =π [ r^{3} - r^{3}/3 + r^{3} - r^{3}/3 ]

V =π [ 2r^{3} - 2r^{3} / 3 ]

**V = 4/3π r ^{3}**

In the following applet, the curve, x^{2} + y^{2} = r^{2}, is rotated around the x-axis, shown in red. The y-axis is shown in green. Please press the *play* button or move the slider. You will see the sphere being formed.

**E.g.4**

Find the volume of the object formed when the curve, y = x^{2} + 2 around the x-axis, between x = 0 and x = 3.

A 'vase' is produced when the curve, **y = x ^{2} + 2** is rotated around the x-axis.

V =π _{0}∫^{3} f(x)^{2} dx

V = π _{0}∫^{3} [ x^{2} + 2 ]^{2} dx

V = π _{0}∫^{3}[x^{4} + 4x^{2} + 4] dx

V =π [ x^{5}/5 + 4x^{3}/3 + 4x ]^{3}_{0}

V =π [ 3^{5}/5 + 4(3)^{3}/3 + 4(3)] - [0^{5}/5 + 4(0)^{0}/3 + 4(0) ]

V =π [96.6]

V = 96.6 π

**V = 96.6π**