5.5 Triple integrals in cylindrical and spherical coordinates  (Page 6/12)

1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.
   For the sphere:
   
   $\begin{array}{ccc}\hfill x^2+y^2+z^2& =\hfill & z\hfill \\ \hfill {\rho }^2& =\hfill & \rho \text{cos}\phi \hfill \\ \hfill \rho & =\hfill & \text{cos}\phi .\hfill \end{array}$
   
   For the cone:
   
   $\begin{array}{ccc}\hfill z& =\hfill & \sqrt{x^2+y^2}\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \sqrt{{\rho }^2{\text{sin}}^2\phi {\text{cos}}^2\varphi +{\rho }^2{\text{sin}}^2\phi {\text{sin}}^2\varphi }\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \sqrt{{\rho }^2{\text{sin}}^2\phi \left({\text{cos}}^2\varphi +{\text{sin}}^2\varphi \right)}\hfill \\ \hfill \rho \text{cos}\phi & =\hfill & \rho \text{sin}\phi \hfill \\ \hfill \text{cos}\phi & =\hfill & \text{sin}\phi \hfill \\ \hfill \phi & =\hfill & \pi \text{/}4.\hfill \end{array}$
   
   Hence the integral for the volume of the solid region $E$ becomes
   
   $V(E)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi =\pi \text{/}4}{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =\text{cos}\phi }{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}.$
2. Consider the $\phi \rho$ -plane. Note that the ranges for $\phi$ and $\rho$ (from part a.) are
   
   $\begin{array}{c}0\le \phi \le \pi \text{/}4\hfill \\ 0\le \rho \le \text{cos}\phi .\hfill \end{array}$
   
   The curve $\rho =\text{cos}\phi$ meets the line $\phi =\pi \text{/}4$ at the point $\left(\pi \text{/}4,\sqrt{2}\text{/}2\right).$ Thus, to change the order of integration, we need to use two pieces:
   
   $\begin{array}{ccccccc}\begin{array}{c}0\le \rho \le \sqrt{2}\text{/}2\hfill \\ 0\le \phi \le \pi \text{/}4\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \sqrt{2}\text{/}2& \le \hfill & \rho \le 1\hfill \\ \hfill 0& \le \hfill & \phi \le {\text{cos}}^{\mathrm{-1}}\rho .\hfill \end{array}\hfill \end{array}$
   
   Hence the integral for the volume of the solid region $E$ becomes
   
   $V(E)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =\sqrt{2}\text{/}2}{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi =\pi \text{/}4}{\int }}{\rho }^2\text{sin}\phi d\phi d\rho d\theta }}}+{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\rho =\sqrt{2}\text{/}2}{\overset{\rho =1}{\int }}{\displaystyle \underset{\phi =0}{\overset{\phi ={\text{cos}}^{\mathrm{-1}}\rho }{\int }}{\rho }^2\text{sin}\phi d\phi d\rho d\theta }}}.$
   
   In each case, the integration results in $V\left(E\right)=\frac{\pi }{8}.$

The ranges of the variables are

The first two inequalities describe the right half of a circle of radius $1.$ Therefore, the ranges for $\theta$ and $r$ are

The limits of $z$ are $r^2\le z\le r,$ hence

The ranges of the variables are

The first two ranges of variables describe a quarter disk in the first quadrant of the $xy$ -plane. Hence the range for $\theta$ is $0\le \theta \le \frac{\pi }{2}.$

The lower bound $z=\sqrt{x^2+y^2}$ is the upper half of a cone and the upper bound $z=\sqrt{18-x^2-y^2}$ is the upper half of a sphere. Therefore, we have $0\le \rho \le \sqrt{18},$ which is $0\le \rho \le 3\sqrt{2}.$

For the ranges of $\phi ,$ we need to find where the cone and the sphere intersect, so solve the equation

This gives

Putting this together, we obtain