Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 13

$$V = 9\pi $$

$$\eqalign{ & {\text{We have }}xy = 4,\,\,\,x + y = 5 \cr & {\text{ }}x = \frac{4}{y},{\text{ and }}x = 5 - y \cr & {\text{Calculate the intersection points Let }}x = x \cr & \frac{4}{y} = 5 - y \cr & 4 = 5y - {y^2} \cr & {y^2} - 5y + 4y = 0 \cr & {y_1} = 1{\text{ and }}{y_2} = 4 \cr & 5 - y \geqslant \frac{4}{y}{\text{ on the interval }}1 \leqslant y \leqslant 4 \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_c^d {2\pi y\left[ {f\left( y \right) - g\left( y \right)} \right]dy} \cr & V = \int_1^4 {2\pi y\left( {5 - y - \frac{4}{y}} \right)dy} \cr & V = 2\pi \int_1^4 {\left( {5y - {y^2} - 4} \right)dy} \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{5}{2}{y^2} - \frac{1}{3}{y^3} - 4y} \right]_1^4 \cr & V = 2\pi \left[ {\frac{5}{2}{{\left( 4 \right)}^2} - \frac{1}{3}{{\left( 4 \right)}^3} - 4\left( 4 \right)} \right] - 2\pi \left[ {\frac{5}{2}{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^3} - 4\left( 1 \right)} \right] \cr & V = 2\pi \left( {\frac{8}{3}} \right) - 2\pi \left( { - \frac{{11}}{6}} \right) \cr & V = 9\pi \cr} $$