Definite Integrals for Calculating Surface Area

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).

Surface area of the rotating object of the area rotated about the X axis

If $y = f\left( x \right)$ is a smooth curve and $f\left( x \right) \ge 0$ in the interval $a \le x \le b$, then the surface area of the object the resulting rotation of the $y = f\left( x \right)$ curve between $x=a$ and $x=b$ about the $X$ axis is determined by the following formula:

\[\boxed{L = 2\pi \int\limits_a^b {f\left( x \right)} .\sqrt {1 + {{\left[ {f'\left( x \right)} \right ]}^2}} dx}\]

Surface area of the rotating object of the area rotated about the Y axis

If $x = g\left( y \right)$ is a smooth curve and $g\left( y \right) \ge 0$ in the interval $c \le y \le d$, then the surface area of the object the resulting rotation of the $x = g\left( y \right)$ curve between $y=c$ and $y=d$ about the $Y$ axis is determined by the following formula:

\[\boxed{L = 2\pi \int\limits_c^d {g\left( y \right)} .\sqrt {1 + {{\left[ {g'\left( y \right)} \right ]}^2}} dy}\]

Examples

Calculate the surface area (cover area) of the following rotating object:

1. The rotating object obtained from the curve $y=4$ rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=3$
2. The rotating object obtained from the curve $y = 2x$ rotated ${360^ \circ }$ about the Y-axis from $y=0$ to $y=2$
3. The rotating object obtained from the curve $y = \sqrt x $ rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=4$

Answer

1. The rotating object obtained from the $y=4$ curve rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=3$

$\begin{array}{l} f\left( x \right) = y = 4\\ f'\left( x \right) = 0\\ a = 0\\ b = 3\\ L = 2\pi \int\limits_a^b {f\left( x \right)} .\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx\\ = 2\pi \int\limits_0^3 {4.\sqrt {1 + {0^2}} } dx\\ = 2\pi \int\limits_0^3 {4.\sqrt 1 } dx\\ = 2\pi \int\limits_0^3 4 dx\\ = 8\pi \left[ x \right]_0^3\\ = 8\pi \left( {3 - 0} \right)\\ = 8\pi \left( 3 \right)\\ = 24\pi \end{array}$

So, the surface area of the rotating object obtained from the curve $y=4$ rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=3$ is $24\pi $ units wide.

2. The rotating object obtained from the curve $y = 2x$ rotated ${360^ \circ }$ about the Y-axis from $y=0$ to $y=2$

$\begin{array}{l} g\left( y \right) = x = \frac{1}{2}y\\ g'\left( y \right) = \frac{1}{2}\\ c = 0\\ d = 2\\ L = 2\pi \int\limits_c^d {g\left( y \right)} .\sqrt {1 + {{\left[ {g'\left( y \right)} \right]}^2}} dy\\ = 2\pi \int\limits_0^2 {\left( {\frac{1}{2}y} \right).\sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2}} } dy\\ = 2\pi \int\limits_0^2 {\left( {\frac{1}{2}y} \right).\sqrt {1 + \left( {\frac{1}{4}} \right)} } dy\\ = \pi \int\limits_0^2 {y.\sqrt {\frac{5}{4}} } dy\\ = \frac{1}{2}\pi \int\limits_0^2 {y.\sqrt 5 } dy\\ = \frac{1}{2}\pi \sqrt 5 \int\limits_0^2 y dy\\ = \frac{1}{2}\pi \sqrt 5 \left[ {\frac{1}{2}{y^2}} \right]_0^2\\ = \frac{1}{2}\pi \sqrt 5 \left( {\frac{1}{2}{{.2}^2} - \frac{1}{2}{{.0}^2}} \right)\\ = \frac{1}{2}\pi \sqrt 5 \left( 2 \right)\\ = \pi \sqrt 5 \end{array}$

So, the surface area of the rotating object obtained from the curve $y = 2x$ rotated ${360^ \circ }$ about the Y axis from $y=0$ to $y=2$ is $\pi \sqrt 5 $ unit area.

3. The rotating object obtained from the curve $y = \sqrt x $ rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=4$

$\begin{array}{l} f\left( x \right) = y = \sqrt x \\ f'\left( x \right) = \frac{1}{{2\sqrt x }}\\ a = 0\\ b = 4\\ L = 2\pi \int\limits_a^b {f\left( x \right)} .\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx\\ = 2\pi \int\limits_0^4 {\sqrt x } .\sqrt {1 + {{\left[ {\frac{1}{{2\sqrt x }}} \right]}^2}} dx\\ = 2\pi \int\limits_0^4 {\sqrt x } .\sqrt {1 + \frac{1}{{4x}}} dx\\ = 2\pi \int\limits_0^4 {\sqrt {x + \frac{1}{4}} dx} \\ = 2\pi \left[ {\frac{2}{3}\left( {x + \frac{1}{4}} \right)\sqrt {x + \frac{1}{4}} } \right]_0^4\\ = \frac{4}{3}\pi \left[ {\left( {\left( {4 + \frac{1}{4}} \right)\sqrt {4 + \frac{1}{4}} } \right) - \left( {\left( {0 + \frac{1}{4}} \right)\sqrt {0 + \frac{1}{4}} } \right)} \right]\\ = \frac{4}{3}\pi \left[ {\left( {\frac{{17}}{4}\sqrt {\frac{{17}}{4}} } \right) - \left( {\left( {\frac{1}{4}} \right)\sqrt {\frac{1}{4}} } \right)} \right]\\ = \frac{4}{3}\pi \left[ {\left( {\frac{{17}}{4}.\frac{1}{2}\sqrt {17} } \right) - \left( {\frac{1}{4}.\frac{1}{2}} \right)} \right]\\ = \frac{4}{3}\pi \left[ {\left( {\frac{{17}}{8}\sqrt {17} } \right) - \left( {\frac{1}{8}} \right)} \right]\\ = \frac{4}{3}\pi .\frac{1}{8}\left( {17\sqrt {17} - 1} \right)\\ = \frac{1}{6}\left( {17\sqrt {17} - 1} \right)\pi \end{array}$

So, the surface area of the rotating object obtained from the curve $y = \sqrt x $ rotated ${360^ \circ }$ about the X axis from $x=0$ to $x=4$ is $\frac {1}{6}\left( {17\sqrt {17} - 1} \right)\pi $ unit area.