Chapter 15 - Multiple Integrals - 15.7 Triple-Integrals in Cylindrical Coordinates - 15.7 Exercises - Page 1083: 17

$384 \pi$

Here, we have $\iiint_E\sqrt{x^2+y^2} dV=\int_0^{2\pi} \int_{-5}^{4}\int_0^{4} r^2 dr dz d\theta$ $=\int_0^{2\pi} d\theta \cdot \int_{-5}^{4} dz \cdot \int_0^{4} r^2 dr$ $=[\theta]_0^{2\pi} [z]_{-5}^{4} \cdot [\dfrac{r^3}{3}]_0^{4}$ $=18 \pi[\dfrac{1}{3}(4)^3-\dfrac{1}{3}(0)^3]$ $=384 \pi$