Triple integrals in spherical coordinates.

  

 

  

Figure 14: The spherical coordinates of a point.

 

Example 7.20   Let $\mathcal{C}$be the cone whose equation is $z=\sqrt{x^2+y^2}$. By substitution from 2.6, we have:
$$\begin{align*}\rho \cos \phi & = \sqrt{ (r \cos \theta )^2 + ( r \sin \theta )^2... ...2 } \\ \rho \cos \phi &= \rho \sin \phi \\ \cos \phi &= \sin \phi \end{align*}$$
i.e., an equation for $\mathcal{C}$in cylindrical coordinates is the following:

$$\phi = \frac {\pi}{4}.$$

 

  

Figure 15: A cone.

 

Example 7.21   See Fig. 16.

• If R is a positive real number, $\rho = R$is the equation of a sphere, whose center is at the origin.
• If $\alpha$is a real number, $\Phi = \alpha$is the equation of a cone, whose vertex is at the origin, and which makes an opening angle of $\alpha$with the positive z-axis.
• If $\alpha$is a real number, $\theta = \alpha$is the equation of a half-plane, whose border is the z-axis, and which makes an angle of $\alpha$with the positive x-axis.

 

  

Figure 16: Simple surfaces in cylindrical coordinates.

 

Example 7.22   Compute the volume of the solid cut from the sphere whose equation is $\rho = 2$by the cone $\theta = \frac {\pi}{4}$.

 

  

Figure 17: An ice cream.

The domain of integration is given by the conditions:

$$0 \leq \rho \leq 2 \qquad ; \qquad 0 \leq \Phi \leq \frac {\pi}{4} \qquad ; \qquad 0 \leq \theta \leq 2 \pi$$

The volume is:
$$\begin{align*}V &= \int_0^{2 \pi} \int_0^{\pi / 4} \int_0^2 \rho^2 \; \sin \Phi ... ... 83 \right) \; d \theta \\ \quad &= \frac {8 \pi}{3} (\sqrt{2}+1). \end{align*}$$