Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 514: 87

$$\eqalign{ & \left. a \right)\pi \left( {1 - {e^{ - 1}}} \right){\text{ cubic units}} \cr & \left. b \right)b = 0.7432 \cr} $$

$$\eqalign{ & {\text{Let the functions }} \cr & y = {e^{ - {x^2}}},\,y = 0{\text{ and }}x = 0 \cr & {\text{The volume by the shell me of the graph shown below revolved}} \cr & {\text{about the }}x{\text{ - axis is given by}} \cr & V = 2\pi \int_a^b {xf\left( x \right)} dx \cr & V = 2\pi \int_0^b {x{e^{ - {x^2}}}} dx \cr & {\text{Rewrite the integrand}} \cr & V = \pi \int_b^0 {\left( { - 2x} \right){e^{ - {x^2}}}} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {{e^{ - {x^2}}}} \right]_b^0 \cr & V = \pi \left[ {{e^{ - {0^2}}} - {e^{ - {b^2}}}} \right] \cr & V = \pi \left( {1 - {e^{ - {b^2}}}} \right) \cr & \cr & \left. a \right){\text{Find the volume when }}b = 1 \cr & V = \pi \left( {1 - {e^{ - {{\left( 1 \right)}^2}}}} \right) \cr & V = \pi \left( {1 - {e^{ - 1}}} \right){\text{ cubic units}} \cr & \cr & \left. b \right){\text{Find }}b{\text{ such as that the volume generated is }}\frac{4}{3} \cr & V = \pi \left( {1 - {e^{ - {b^2}}}} \right) \cr & {\text{Let }}b = \frac{4}{3} \cr & \frac{4}{3} = \pi \left( {1 - {e^{ - {b^2}}}} \right) \cr & {\text{Solve for }}b \cr & \frac{4}{{3\pi }} = 1 - {e^{ - {b^2}}} \cr & {e^{ - {b^2}}} = 1 - \frac{4}{{3\pi }} \cr & \ln \left( {{e^{ - {b^2}}}} \right) = \ln \left( {1 - \frac{4}{{3\pi }}} \right) \cr & - {b^2} = \ln \left( {\frac{{3\pi - 4}}{{3\pi }}} \right) \cr & {b^2} = \ln \left( {\frac{{3\pi }}{{3\pi - 4}}} \right) \cr & b = \pm \sqrt {\ln \left( {\frac{{3\pi }}{{3\pi - 4}}} \right)} \cr & {\text{Where }}b > 0,{\text{ then}} \cr & b = \sqrt {\ln \left( {\frac{{3\pi }}{{3\pi - 4}}} \right)} \approx 0.7432 \cr} $$