Section 4.6 Spherical Integral

4.6 Spherical Integral

Recall the Spherical Coordinates system denotes a point in a 3-dimensional space via $(\rho ,\theta ,\phi )$ where $\rho$ is the distance between the point and origin, $\theta$ is the angle between the point and positive $x$-axis, and $\phi$ is the angle between the point and positive $z$-axis. Hence we have relationships, 

$$x^2+y^2+z^2={\rho }^2,x=\rho \text{sin}(\phi )\text{cos}(\theta ),y=\rho \text{sin}(\phi )\text{sin}(\theta ),z=\rho \text{cos}(\phi ).$$

Definition: Triple Integrals

Consider the spherical box (expressed in spherical coordinates)  $$B=\{(\rho ,\theta ,\phi )|a\le \rho \le b,\alpha \le \theta \le \beta ,c\le \phi \le d\}.$$

If the function $f(\rho ,\theta ,\phi )$ is continuous on $B$ and if $({\rho }_{ijk}^{\ast },{\theta }_{ijk}^{\ast },{\phi }_{ijk}^{\ast })$ is any sample point in the spherical $B_{ijk}=[{\rho }_{i-1},{\rho }_i]\times [{\theta }_{j-1},{\theta }_j]\times [{\phi }_{k-1},{\phi }_k]$, then we can define the triple integral in spherical coordinates as the limit of a triple Riemann sum, provided the following limit exist

$$\underset{l,m,n\to \mathrm{\infty }}{lim}\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}f({\rho }_{ijk}^{\ast },{\theta }_{ijk}^{\ast },{\phi }_{ijk}^{\ast })({\rho }_{ijk}^{\ast }{)}^2\text{sin}(\phi )\mathrm{△}\rho \mathrm{△}\theta \mathrm{△}\phi$$

 if the limit exists.

Theorem: Fubini’ s Theorem in Cylindrical Coordinates

Suppose that $g(x,y,z)$ is continuous on a box B, which when described in spherical coordinates looks like $B=\{(\rho ,\theta ,\phi )|a\le \rho \le b,\alpha \le \theta \le \beta ,c\le \phi \le d\}$.

Then $g(x,y,z)=g(\rho \text{sin}(\phi )\text{cos}(\theta ),\rho \text{sin}(\phi )\text{sin}(\theta ),\rho \text{cos}(\phi ))=f(\rho ,\theta ,\phi )$ and

$$\int \int {\int }_{B}g(x,y,z)dV={\int }_c^d{\int }_{\alpha }^{\beta }{\int }_a^{b}f(\rho ,\theta ,\phi ){\rho }^2\text{sin}(\phi )d\rho d\theta d\phi .$$

This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.

Example 1: Evaluating the triple integral ${\int }_0^{\pi }{\int }_0^{\frac{\pi }{2}}{\int }_0^1{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Exercise 1: Evaluating the triple integral ${\int }_0^{\frac{\pi }{2}}{\int }_0^{\pi }{\int }_0^2{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Remark (Theorem): As before, we can do the integration over a general box, not just a spherical one. 

Example 2: Let $E$ be the region bounded above by the cone $z=-\sqrt{x^2+y^2}$ and below by the sphere $z^2+x^2+y^2=1$ . Set up a triple integral in spherical coordinates to find the volume of the region.

Exercise 2: Let $E$ be the region bounded below by the cone $z=\sqrt{3x^2+3y^2}$ and above by the sphere $z^2+x^2+y^2=4$ . Set up a triple integral in spherical coordinates to find the volume of the region.

Example 3: Sketch the solid whose volume is given by the integral and evaluate the integral ${\int }_0^{\frac{\pi }{6}}{\int }_0^{\frac{\pi }{3}}{\int }_0^1{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Exercise 3: Sketch the solid whose volume is given by the integral and evaluate the integral ${\int }_0^{\frac{\pi }{4}}{\int }_0^{\pi }{\int }_0^2{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Example 4: Sketch the solid whose volume is given by the integral and evaluate the integral ${\int }_0^{\frac{\pi }{6}}{\int }_0^{\frac{\pi }{2}}{\int }_0^{2\text{sec}(\phi )}{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Exercise 4: Sketch the solid whose volume is given by the integral and evaluate the integral ${\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{\pi }{3}}{\int }_0^{\text{sec}(\phi )}{\rho }^2\text{sin}(\phi )d\rho d\theta d\phi$.

Example 5: Sketch the solid. Evaluating the triple integral $\int \int {\int }_{E}xdV$ where $E$ is the solid of hemisphere $x^2+y^2+z^2\le 4$, $x\ge 0$.

Exercise 5: Sketch the solid. Evaluating the triple integral $\int \int {\int }_{E}ydV$ where $E$ is the solid of hemisphere $x^2+y^2+z^2\le 9$, $y\ge 0$.

Example 6: Sketch the solid and find its volume. The solid lies inside both of sphere $x^2+y^2+z^2=4$, and the cone $\sqrt{x^2+y^2}=z$. 

Exercise 6: Sketch the solid and find its volume. The solid lies inside sphere $x^2+y^2+z^2=9$, and outside the cone $\sqrt{x^2+y^2}=z$.

Example 7: Sketch the solid and find its volume. The solid lies above the cone $\phi =\frac{\pi }{3}$ and below the sphere $\rho =4\text{cos}(\phi )$

Example 8: Change the integration into spherical coordinates

$${\int }_0^1{\int }_0^{\sqrt{1-x^2}}{\int }_{\sqrt{x^2+y^2}}^{\sqrt{2-x^2-y^2}}xzdzdydx$$

Group work:

1. Sketch the solid and find its volume. The solid lies above the cone $\phi =\frac{\pi }{4}$ and below the sphere $\rho =6\text{cos}(\phi )$.

2. Change the integration into spherical coordinates

$${\int }_{-4}^4{\int }_0^{\sqrt{16-x^2}}{\int }_{-\sqrt{16-x^2-y^2}}^{\sqrt{16-x^2-y^2}}{x}^2+y^{2}dzdydx.$$

3. Sketch the solid and find its volume. The solid lies inside between the spheres $x^2+y^2+z^2=4$ and $x^2+y^2+z^2=1$ and above the cone $z=\sqrt{x^2+y^2}$. 

4. Sketch the solid and find its volume. The solid is obtained by cutting out a smaller wedge from a sphere of radius 4 by two planes intersect along a diameter at an angle of $\frac{\pi }{3}$.