Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 321: 12

$15$

Area $=b^2=(9/25) \times (5-z)^2$ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{5} (9/25) \times (5-z)^2 dz$ Now, $V= (-\dfrac{3}{25})(5-z)^3]_{0}^{5}$ or, $= (-\dfrac{3}{25}) (0-125)$ or, $=15$