Chapter 16 - Multiple Integration - 16.4 Integration in Polar, Cylindrical, and Spherical Coordinates - Exercises - Page 881: 42

The volume of the region is $\frac{5}{3}\pi $.

We have the region ${\cal W}$ bounded below by the plane $z=1$ and above by the sphere ${x^2} + {y^2} + {z^2} = 4$. Thus, $z$ varies from $z=1$ to $z = \sqrt {4 - {x^2} - {y^2}} $. In cylindrical coordinates we have $1 \le z \le \sqrt {4 - {r^2}} $. We find the intersection of the plane and the sphere by substituting $z=1$ in ${x^2} + {y^2} + {z^2} = 4$: ${x^2} + {y^2} + 1 = 4$ ${x^2} + {y^2} = 3$ So, the intersection is a circle of radius $\sqrt 3 $. Thus, the projection of ${\cal W}$ onto the $xy$-plane is a disk ${x^2} + {y^2} \le 3$; in cylindrical coordinates is described by ${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 3 ,0 \le \theta \le 2\pi } \right\}$ Thus, the region description of ${\cal W}$: ${\cal W} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 3 ,0 \le \theta \le 2\pi ,1 \le z \le \sqrt {4 - {r^2}} } \right\}$ Evaluate the volume of ${\cal W}$: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 3 } \mathop \smallint \limits_{z = 1}^{\sqrt {4 - {r^2}} } r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 3 } r\left( {z|_1^{\sqrt {4 - {r^2}} }} \right){\rm{d}}r{\rm{d}}\theta $ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 3 } \left( {r\sqrt {4 - {r^2}} - r} \right){\rm{d}}r{\rm{d}}\theta $ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( { - \frac{1}{3}{{\left( {4 - {r^2}} \right)}^{3/2}} - \frac{1}{2}{r^2}} \right)|_0^{\sqrt 3 }} \right){\rm{d}}\theta $ $ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - \frac{1}{3} - \frac{3}{2} + \frac{8}{3}} \right){\rm{d}}\theta $ $ = \frac{5}{6}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{5}{3}\pi $ Thus, the volume of the region is $\frac{5}{3}\pi $.