# Chapter 14 - Section 14.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises - Page 803: 2

$$\dfrac{9\pi(8\sqrt{2}-7)}{2}$$

#### Work Step by Step

Our aim is to integrate the integral as follows: $$\int^{2\pi}_0 \int^3_0 \int^{\sqrt{18-r^2}}_{r^2/3} dz \space r \space dr \space d \theta = \int^{2\pi}_0 \int^{3}_0[r \times (18-r^2)^{1/2}-\dfrac{1}{3} r^3] \space dr \space d\theta \\=\int^{2\pi}_0 [\dfrac{-1}{3} \times (18-r^2)+\dfrac{1}{12} r^4]^{3}_0 d\theta \\=\dfrac{9\pi(8\sqrt{2}-7)}{2}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.