Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 515: 61

$$\eqalign{ & \left( {\text{a}} \right)V = \frac{{14}}{3}\pi \cr & \left( {\text{b}} \right)V = \left( {\frac{{2\sqrt 5 + 2}}{3}} \right)\pi \cr} $$

$$\eqalign{ & f\left( x \right) = \sqrt {{x^2} + 1} {\text{ and the }}x{\text{ - axis on the interval }}\left[ {0,2} \right] \cr & \cr & \left( {\text{a}} \right){\text{Calculate the volume using the disk method about the x - axis}} \cr & V = \int_a^b {\pi f{{\left( x \right)}^2}dx} \cr & V = \pi \int_0^2 {{{\left( {\sqrt {{x^2} + 1} } \right)}^2}dx} \cr & V = \pi \int_0^2 {\left( {{x^2} + 1} \right)dx} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{3}{x^3} + x} \right]_0^2 \cr & V = \pi \left[ {\frac{1}{3}{{\left( 2 \right)}^3} + 2} \right] - \pi \left[ 0 \right] \cr & V = \pi \left( {\frac{{14}}{3}} \right) \cr & V = \frac{{14}}{3}\pi \cr & \cr & \left( {\text{b}} \right) \cr & f\left( x \right) = \sqrt {{x^2} + 1} \cr & y = \sqrt {{x^2} + 1} \cr & {\text{Solve for }}x \cr & {y^2} = {x^2} + 1 \cr & {x^2} = {y^2} - 1 \cr & x = \sqrt {{y^2} - 1} \cr & x = 0 \Rightarrow y = 1 \cr & x = 2 \Rightarrow y = \sqrt 5 \cr & f\left( y \right) = \sqrt {{y^2} - 1} \cr & {\text{using the disk method about the y - axis}} \cr & V = \int_c^d {\pi f{{\left( y \right)}^2}dx} \cr & V = \int_1^{\sqrt 5 } {\pi {{\left( {\sqrt {{y^2} - 1} } \right)}^2}dy} \cr & V = \pi \int_1^{\sqrt 5 } {\left( {{y^2} - 1} \right)dy} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{3}{y^3} - y} \right]_1^{\sqrt 5 } \cr & V = \pi \left[ {\frac{1}{3}{{\left( {\sqrt 5 } \right)}^3} - \sqrt 5 } \right] - \pi \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - 1} \right] \cr & V = \pi \left[ {\frac{{2\sqrt 5 }}{3}} \right] - \pi \left[ { - \frac{2}{3}} \right] \cr & V = \left( {\frac{{2\sqrt 5 + 2}}{3}} \right)\pi \cr} $$