Chapter 8 - Further Techniques and Applications of Integration - 8.2 Volume and Average Value - 8.2 Exercises - Page 439: 6

$\frac{1685}{12}\pi$

Use the formula given above for the volume, with a=0, b=5 $V=\int^{5}_{0} \pi (\frac{1}{2}x +4)^{2}dx=\int^{5}_{0}\pi(\frac{1}{4}x^{2}+4x+16)dx=\pi [\frac{x^{3}}{12}+2x^{2}+16x]|^{5}_{0}=\frac{1685}{12}\pi$