Definite Integrals for Calculating the Volume of Rotating Objects

Definition of Rotating Object

A rotating object is a space object obtained from the rotation of an area about a certain line (the axis of rotation).

Volume of Rotating Object from Area Rotated About X-Axis

If the area bounded by the curve $y = f\left( x \right)$, the axis $X$, the line $x=a$, and the line $x=b$ are rotated by ${360^ \circ } $ around the $X$ axis, then the volume or content of the rotating object that occurs is determined by the formula:

\[\begin{array}{l} \boxed{V = \pi \int\limits_a^b {{y^2}} dx}\\ or\\ \boxed{V = \pi \int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}} dx} \end{array}\]

Volume of Rotating Object from Area Rotated About Y Axis

If the area bounded by the curve $x = g\left( y \right)$, the axis $Y$, the line $y=c$, and the line $y=d$ are rotated by ${360^ \circ } $ around the $Y$ axis, then the volume or content of the rotating object that occurs is determined by the formula:

\[\begin{array}{l} \boxed{V = \pi \int\limits_c^d {{x^2}} dy}\\ or\\ \boxed{V = \pi \int\limits_c^d {{{\left[ {g\left( y \right)} \right]}^2}} dy} \end{array}\]

The Volume of the Rotating Body of the Region Between Two Curves Rotated About the X-Axis

If the area bounded by the curve ${y_1} = f\left( x \right)$, curve ${y_2} = g\left( x \right)$, line $x=a$, and line $ x=b$ rotated as far as ${360^ \circ }$ around the axis $X$, then the volume or content of the rotating object that occurs is determined by the formula:

\[\begin{array}{l} \boxed{V = \pi \int\limits_a^b {\left( {y_1^2 - y_2^2} \right)} dx}\\ or\\ \boxed{V = \pi \int\limits_a^b {\left[ {{f^2}\left( x \right) - {g^2}\left( x \right)} \right]} dx} \end{array}\]

The Volume of the Rotating Body of the Region Between Two Curves Rotated About the Y Axis

If the area bounded by the curve ${x_1} = f\left( y \right)$, curve ${x_2} = g\left( y \right)$, line $y=c$, and line $ y=d$ rotated as far as ${360^ \circ }$ around the axis $Y$, then the volume or content of the rotating object that occurs is determined by the formula:

\[\begin{array}{l} \boxed{V = \pi \int\limits_c^d {\left( {x_1^2 - x_2^2} \right)} dy}\\ or\\ \boxed{V = \pi \int\limits_c^d {\left[ {{f^2}\left( y \right) - {g^2}\left( y \right)} \right]} dy} \end{array}\]

Examples

1. Find the volume of the rotating object in the following shaded regions if it is rotated ${360^ \circ }$ around the $X$ axis!


2. Find the volume of the rotating object in the following shaded areas if it is rotated ${360^ \circ }$ around the $Y$ axis!

Answer

1. The shaded area is delimited by the $y=2$ line, the $X$ axis, the $x=0$ line, and $x=3$. If the shaded area is rotated a distance of ${360^ \circ }$ around the $X$ axis, the volume of the rotating object is defined as follows:
$\begin{array}{l} f\left( x \right) = y = 2\\ a = 0\\ b = 3\\ V &= \pi \int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}} dx\\ &= \pi \int\limits_0^3 {{{\left[ 2 \right]}^2}} dx\\ &= \pi \int\limits_0^3 4 dx\\ &= \pi \left[ {4x} \right]_0^3\\ &= \pi \left( {4.3 - 4.0} \right)\\ &= 12\pi \end{array}$

So, the volume of the rotating object that occurs is $12\pi $ unit volume.

2. The shaded area is bounded by the curve $f\left( x \right) = {y_1} = - x + 2$ and the curve $g\left( x \right) = {y_2} = {x^2}$ with $f\left( x \right) \ge g\left( x \right)$ in the shaded area. Before determining the volume, first determine the lower and upper limits of the integration as follows:
$\begin{array}{l} f\left( x \right) = g\left( x \right)\\ \Leftrightarrow - x + 2 = {x^2}\\ \Leftrightarrow {x^2} + x - 2 = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0\\ x + 2 = 0 \Rightarrow x = - 2\\ x - 1 = 0 \Rightarrow x = 1\\ a = - 2\\ b = 1 \end{array}$

If the shaded region is rotated ${360^ \circ }$ around the $X$ axis, the volume of the rotating object is defined as follows:

$\begin{array}{l} V = \pi \int\limits_a^b {\left[ {{f^2}\left( x \right) - {g^2}\left( x \right)} \right]} dx\\ = \pi \int\limits_{ - 2}^1 {\left[ {{{\left( { - x + 2} \right)}^2} - {{\left( {{x^2}} \right)}^2}} \right]} dx\\ = \pi \int\limits_{ - 2}^1 {\left[ {{x^2} - 4x + 4 - {x^4}} \right]} dx\\ = \pi \left[ {\frac{1}{3}{x^3} - 2{x^2} + 4x - \frac{1}{5}{x^5}} \right]_{ - 2}^1\\ = \pi \left[ {\left( {\frac{1}{3}{{.1}^3} - {{2.1}^2} + 4.1 - \frac{1}{5}{{.1}^5}} \right) - \left( {\frac{1}{3}.{{\left( { - 2} \right)}^3} - 2.{{\left( { - 2} \right)}^2} + 4.\left( { - 2} \right) - \frac{1}{5}.{{\left( { - 2} \right)}^5}} \right)} \right]\\ = \pi \left[ {\left( {\frac{1}{3} - 2 + 4 - \frac{1}{5}} \right) - \left( { - \frac{8}{3} - 8 - 8 + \frac{{32}}{5}} \right)} \right]\\ = \frac{{72}}{5}\pi \end{array}$

So, the volume of the rotating object that occurs is $\frac{{72}}{5}\pi $ unit volume.

1. The shaded area is delimited by the line $y = 3x - 2$, the axis $Y$, the line $y=1$, and $y=2$. If the shaded region is rotated a distance of ${360^ \circ }$ around the $Y$ axis, the volume of the rotating object that occurs is defined as follows:
$\begin{array}{l} y = 3x - 2\\ \Leftrightarrow 3x = y + 2\\ \Leftrightarrow x = \frac{1}{3}y + \frac{2}{3}\\ g\left( y \right) = x = \frac{1}{3}y + \frac{2}{3}\\ c = 1\\ d = 2\\ V = \pi \int\limits_1^2 {{{\left[ {g\left( y \right)} \right]}^2}dy} \\ = \pi \int\limits_1^2 {{{\left[ {\frac{1}{3}y + \frac{2}{3}} \right]}^2}dy} \\ = \pi \int\limits_1^2 {\left( {\frac{1}{9}{y^2} + \frac{4}{9}y + \frac{4}{9}} \right)dy} \\ = \frac{1}{9}\pi \int\limits_1^2 {\left( {{y^2} + 4y + 4} \right)dy} \\ = \frac{1}{9}\pi \left[ {\frac{1}{3}{y^3} + 2{y^2} + 4y} \right]_1^2\\ = \frac{1}{9}\pi \left[ {\left( {\frac{1}{3}{{.2}^3} + {{2.2}^2} + 4.2} \right) - \left( {\frac{1}{3}{{.1}^3} + {{2.1}^2} + 4.1} \right)} \right]\\ = \frac{1}{9}\pi \left[ {\left( {\frac{8}{3} + 8 + 8} \right) - \left( {\frac{1}{3} + 2 + 4} \right)} \right]\\ = \frac{1}{9}\pi \left( {\frac{8}{3} + 8 + 8 - \frac{1}{3} - 2 - 4} \right)\\ = \frac{{37}}{{27}}\pi \end{array}$

So, the volume of the rotating object that occurs is $\frac{{37}}{{27}}\pi $ unit volume.

2. The shaded area is bounded by the curve $y = {x^2} \Rightarrow f\left( y \right) = \sqrt y $ and the curve $y = \sqrt x \Rightarrow g\left( y \right) = {y^2}$ with $f\left( y \right) \ge g\left( y \right)$ in the shaded region. Before determining the volume, first determine the lower and upper limits of the integration as follows:
$\begin{array}{l} f\left( y \right) = g\left( y \right)\\ \Leftrightarrow \sqrt y = {y^2}\\ \Leftrightarrow y = {y^4}\\ \Leftrightarrow y - {y^4} = 0\\ \Leftrightarrow y\left( {1 - {y^3}} \right) = 0\\ \Leftrightarrow y = 0 \vee 1 - {y^3} = 0\\ \Leftrightarrow y = 0 \vee y = 1\\ c = 0\\ d = 1 \end{array}$

If the shaded area is rotated a distance of ${360^ \circ }$ around the $Y$ axis, the volume of the rotating object that occurs is formulated as follows:

$\begin{array}{l} V = \pi \int\limits_0^1 {\left[ {{f^2}\left( y \right) - {g^2}\left( y \right)} \right]dy} \\ = \pi \int\limits_0^1 {\left[ {{{\left( {\sqrt y } \right)}^2} - {{\left( {{y^2}} \right)}^2}} \right]dy} \\ = \pi \int\limits_0^1 {\left[ {y - {y^4}} \right]dy} \\ = \pi \left[ {\frac{1}{2}{y^2} - \frac{1}{5}{y^5}} \right]_0^1\\ = \pi \left[ {\left( {\frac{1}{2}{{.1}^2} - \frac{1}{5}{{.1}^5}} \right) - \left( {\frac{1}{2}{{.0}^2} - \frac{1}{5}{{.0}^5}} \right)} \right]\\ = \pi \left[ {\left( {\frac{1}{2} - \frac{1}{5}} \right) - \left( 0 \right)} \right]\\ = \frac{3}{{10}}\pi \end{array}$

So, the volume of the rotating object that occurs is $\frac{3}{{10}}\pi $ unit volume.