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Chapter 7: Triple Integration

Section 7.5: Integration in Spherical Coordinates

Example 7.6.3

Use spherical coordinates to integrate the function $f = 1$ over $R$ , the region inside the unit sphere centered at the origin, and outside the cone $z = \sqrt{x^{2} + y^{2}}$ .

(This region was graphed in Example 7.5.5.)

Solution

Mathematical Solution

•	The graph in Figure 7.6.3(a) was obtained in Example 7.5.3.

•	Iteration in the order $d ρ d φ d θ$ leads to

$\int_{0}^{2 π} \int_{π / 4}^{π / 2} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ$ = $\frac{1}{3} \sqrt{2} π$

•	The lower limit of the middle integral is determined by the cone $z = \sqrt{x^{2} + y^{2}}$ ; the upper limit, by recalling that $φ \in [0, π]$ .

use plots in
module()
local p1,p2,p3;
p1:=plot3d([rho,theta,Pi/4],rho=0..1,theta=0..2*Pi,coords=spherical);
p2:=plot3d(1,theta=0..2*Pi,phi=Pi/4..Pi/2,coords=spherical, transparency=.3);
p3:=display(p1,p2,axes=frame,labels=[x,y,z],tickmarks=[3,3,2], orientation=[-10,75,0],lightmodel=none,scaling=constrained);
print(p3);
end module:
end use:

Figure 7.6.3(a) The region $R$

Maple Solution - Interactive

Table 7.6.3(a) provides a solution by a visualization task template.

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Spherical

Evaluate $∭_{R} Ψ (ρ, φ, θ) dv$ and Graph $R$

Volume Element $dv = ρ^{2} \sin (φ) \times$

$d ρ d φ d θ$

$d ρ d θ d φ$

$d φ d ρ d θ$

$d φ d θ d ρ$

$d θ d φ d ρ$

$d θ d ρ d φ$

, where $Ψ =$

$F =$	$G =$	$b =$
$f =$	$g =$	$a =$

Table 7.6.3(a) Solution by visualization task template

In Table 7.6.3(a), constrained scaling has been applied to the graph via its Context Panel.

Table 7.6.3(b) provides a solution by a task template that embodies the MultiInt command from the Student MultivariateCalculus package. It iterates in just the order $d ρ d φ d θ$ , although the command itself will iterate in any of the six possible orders.

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Multiple Integration≻Spherical

Iterated Triple Integral in Spherical Coordinates

( $φ$ = colatitude, measured down from $z$ -axis)

Integrand:

$1$

(1)

Region: $\{ρ_{1} (φ, θ) \leq ρ \leq ρ_{2} (φ, θ), φ_{1} (θ) \leq φ \leq φ_{2} (θ), a \leq θ \leq b\}$

$ρ_{1} (φ, θ)$

$0$

(2)

$ρ_{2} (φ, θ)$

$1$

(3)

$φ_{1} (θ)$

$π / 4$

$\frac{1}{4} π$

(4)

$φ_{2} (θ)$

$π / 2$

$\frac{1}{2} π$

(5)

$a$

$0$

(6)

$b$

$2 π$

(7)

Inert Integral: $d ρ d φ d θ$

>	$Student [MultivariateCalculus] [MultiInt] (, ρ = .., φ = .., θ = .., coordinates = spherical [ρ, φ, θ], output = integral)$

$\int_{0}^{2 π} \int_{\frac{1}{4} π}^{\frac{1}{2} π} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ$

(8)

Value:

>	$Student [MultivariateCalculus] [MultiInt] (, ρ = .., φ = .., θ = .., coordinates = spherical [ρ, φ, θ])$

$\frac{1}{3} \sqrt{2} π$

(9)

Stepwise Evaluation:

>	$Student [MultivariateCalculus] [MultiInt] (, ρ = .., φ = .., θ = .., coordinates = spherical [ρ, φ, θ], output = steps)$

$\begin{array}{c} \int_{0}^{2 π} \int_{\frac{π}{4}}^{\frac{π}{2}} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} \int_{\frac{π}{4}}^{\frac{π}{2}} (\binom{\frac{ρ^{3} \sin (φ)}{3}}{} | \binom{}{ρ = 0 .. 1}) &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\sin (φ)}{3} &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} (\binom{- \frac{\cos (φ)}{3}}{} | \binom{}{φ = \frac{π}{4} .. \frac{π}{2}}) &DifferentialD; θ \\ = & \int_{0}^{2 π} \frac{\sqrt{2}}{6} &DifferentialD; θ \\ = & \binom{\frac{\sqrt{2} θ}{6}}{} | \binom{}{θ = 0 .. 2 π} \end{array}$

$\frac{1}{3} \sqrt{2} π$

(10)

Table 7.6.3(b) Solution by task template that embodies the MultiInt command

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

•	Type the integrand, 1.

•	Context Panel: Student Multivariate Calculus≻Integrate≻Iterated Fill in the fields of the two dialogs shown below.

•	Context Panel: Evaluate Integral

$1$ $\overset{MultiInt}{\to}$ $\int_{0}^{2 π} \int_{\frac{1}{4} π}^{\frac{1}{2} π} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ$ $=$ $\frac{1}{3} \sqrt{2} π$

Table 7.6.3(c) contains a solution from first principles, with the iterated integral being set via the iterated triple-integral template in the Calculus palette.

•	Calculus palette: Iterated triple-integral template (In the integrand, be sure to include the Jacobian for spherical coordinates.)

•	Context Panel: Evaluate and Display Inline

$\int_{0}^{2 π} \int_{π / 4}^{π / 2} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ$ = $\frac{1}{3} \sqrt{2} π$

Table 7.6.3(c) Solution from first principles

Maple Solution - Coded

Initialize

•	Install the Student MultivariateCalculus package.

$with (Student :- MultivariateCalculus) &colon;$

Apply the MultiInt command from the Student MultivariateCalculus package

$MultiInt (1, ρ = 0 .. 1, φ = π / 4 .. π / 2, θ = 0 .. 2 π, coordinates = spherical [ρ, φ, θ], output = integral)$

$\int_{0}^{2 π} \int_{\frac{1}{4} π}^{\frac{1}{2} π} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ$

$MultiInt (1, ρ = 0 .. 1, φ = π / 4 .. π / 2, θ = 0 .. 2 π, coordinates = spherical [ρ, φ, θ])$

$\frac{1}{3} \sqrt{2} π$

$MultiInt (1, ρ = 0 .. 1, φ = π / 4 .. π / 2, θ = 0 .. 2 π, coordinates = spherical [ρ, φ, θ], output = steps)$

Solution via the top-level Int and int commands

$Int (ρ^{2} \sin (φ), ρ = 0 .. 1, φ = π / 4 .. π / 2, θ = 0 .. 2 π) = int (ρ^{2} \sin (φ), ρ = 0 .. 1, φ = π / 4 .. π / 2, θ = 0 .. 2 π)$

$\int_{0}^{2 π} \int_{\frac{1}{4} π}^{\frac{1}{2} π} \int_{0}^{1} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ = \frac{1}{3} \sqrt{2} π$

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