Chapter 6 - Applications of the Integral - 6.2 Setting Up Integrals: Volume, Density, Average Value - Exercises - Page 297: 17

$V$ = $96\pi$

$h_1$ $\frac{6}{12}$ = $\frac{h_1}{12-y}$ $h_1$ = $\frac{1}{2}(12-y)$ $h_2$ $\frac{4}{12}$ = $\frac{h_2}{12-y}$ $h_2$ = $\frac{1}{3}(12-y)$ $A$ = $\frac{\pi}{6}(12-y)^{2}$ $V$ = $\int_0^{12}\frac{\pi}{6}(12-y)^{2}dy$ $V$ = $-\frac{\pi}{18}(12-y)^{3}|_0^{12}$ $V$ = $96\pi$