# Integration: Volume of a Revolution

The figure above shows the function $y={x}^{2}$ rotated around the y-axis, which creates a solid with a contour following that of the function. Integration allows us to calculate this volume. For a solid of revolution rotated around the y-axis, the general definition is:

$V=\pi \int _{ a }^{ b }{ { x }^{ 2 }\quad dy }$

where x is a function of y, from boundaries a to b.

Meanwhile, the solid of revolution rotated around the x-axis, the general definition is:

$V=\pi \int _{ a }^{ b }{ { y }^{ 2 }\quad dx }$

where y is a function of x, from boundaries a to b.

Explore the revolution of solids using the Geogebra app below.

### Examples

1. Find the volume of the solid of revolution formed when the curve ${x}^{2}+{y}^{2}=9$ is rotated about the x-axis between x=1 and x=3.

Notice that the solid is rotated about the x-axis. Therefore, the boundary is in dx. First, change the subject to ${y}^{2}$

${ x }^{ 2 }+{ y }^{ 2 }=9\\ { y }^{ 2 }=9-{ x }^{ 2 }\\ \\ V=\pi \int _{ a }^{ b }{ { y }^{ 2 }\quad dx } \\ \\ V=\pi \int _{ 1 }^{ 3 }{ { 9-{ x }^{ 2 } }\quad dx } \\ \\ V=\pi { \left[ 9x-\cfrac { { x }^{ 3 } }{ 3 } \right] }_{ 1 }^{ 3 }\\ \\ V=\pi \left[ \left( 27-9 \right) -(9-\cfrac { 1 }{ 3 } ) \right] \\ \\ V=\cfrac { 28\pi }{ 3 } \quad cubic\quad units$

2. Find the volume of the solid formed when the curve $y={x}^{2}-1$ is rotated about the y-axis from y=-1 to y=3.

Notice that the volume is rotated about the y-axis. Therefore, the boundary is in dy. First, change the subject to ${x}^{2}$.
$y={ x }^{ 2 }-1\\ { x }^{ 2 }=y+1\\ \\ V=\pi \int _{ a }^{ b }{ { x }^{ 2 }\quad dy } \\ \\ V=\pi \int _{ -1 }^{ 3 }{ y+1\quad dy } \\ \\ V={ \left[ \cfrac { { y }^{ 2 } }{ 2 } +y \right] }_{ -1 }^{ 3 }\\ \\ V=\left[ \left( \cfrac { 9 }{ 2 } +3 \right) -\left( \cfrac { 1 }{ 2 } -1 \right) \right] \\ \\ V=8\pi \quad cubic\quad units$