Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 357: 60

$Volume= \pi (\dfrac{36}{5})$ and $\space Weight =192 \space g$

We need to integrate the integral to compute the volume. We have: $$Volume = \int_{0}^{6} (36-x^2) \dfrac{x^2}{144} dx \times \pi \\= \dfrac{ \pi }{144} [12x^3 - \dfrac{x^{5}}{5}]_{0}^{6} \\= \pi (\dfrac{36}{5})$$ Now, $$Weight =\pi (\dfrac{36}{5}) \approx 192 \space g$$