3.6 Integration as the Summation of Volumes

The volume of the solid generated when the region enclosed by the curve y = f(x), the x-axis, the line x = a and the line x = b is revolved through 360° about the x-axis is given by

$V_x=\pi {\displaystyle {\int }_a^b{y}^{2}dx}$

(2).



The volume of the solid generated when the region enclosed by the curve x = f(y), the y-axis, the line y = a and the line y = b is revolved through 360° about the y-axis is given by

$$V_y=\pi {\displaystyle {\int }_a^b{x}^{2}dy}$$

Example 1:


Solution:
Volume generated, V_x
$\begin{array}{l}{V}_x=\pi {\displaystyle {\int }_a^b{y}^{2}dx}\\ V_x=\pi {\displaystyle {\int }_2^4{\left(3x-\frac{8}{x}\right)}^{2}dx}\\ V_x=\pi {\displaystyle {\int }_2^4\left(3x-\frac{8}{x}\right)\left(3x-\frac{8}{x}\right)dx}\\ V_x=\pi {\displaystyle {\int }_2^4\left(9x^2-48+\frac{64}{x^2}\right)dx}\\ V_x=\pi {\left[\frac{9x^3}{3}-48x+\frac{64x^{-1}}{-1}\right]}_2^4\\ V_x=\pi {\left[3x^3-48x-\frac{64}{x}\right]}_2^4\\ V_x=\pi \left[\left(3{(4)}^3-48(4)-\frac{64}{4}\right)-\left(3{(2)}^3-48(2)-\frac{64}{2}\right)\right]\\ V_x=\pi \left(-16+104\right)\\ V_x=88\pi uni{t}^3\end{array}$

Example 2:
Solution:
Volume generated, V_y
$\begin{array}{l}{V}_y=\pi {\displaystyle {\int }_a^b{x}^{2}dy}\\ V_y=\pi {\displaystyle {\int }_1^2{\left(\frac{2}{y}\right)}^{2}dy}\\ V_y=\pi {\displaystyle {\int }_1^2\left(\frac{4}{y^2}\right)dy}\\ V_y=\pi {\displaystyle {\int }_1^{2}4y^{-2}dy}\\ V_y=\pi {\left[\frac{4y^{-1}}{-1}\right]}_1^2=\pi {\left[-\frac{4}{y}\right]}_1^2\\ V_y=\pi \left[\left(-\frac{4}{2}\right)-\left(-\frac{4}{1}\right)\right]\\ V_y=2\pi uni{t}^3\end{array}$