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Chapter 7: Triple Integration

Section 7.5: Integration in Spherical Coordinates

Example 7.6.2

Use spherical coordinates to integrate the function $f = 1$ over $R$ , the region above the cone $z = \sqrt{x^{2} + y^{2}}$ but below the unit sphere that is centered at $(0, 0, 1)$ .

(This region was graphed in Example 7.5.4.)

(See Example 7.4.11 where this integral is evaluated in cylindrical coordinates.)

Solution

Mathematical Solution

•	The graph in Figure 7.6.2(a) was obtained in Example 7.5.4.

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

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

•	The astute reader will note that the value of the integral, namely $π$ , is the volume of the region $R$ . This region consists of a hemisphere of radius 1 (having volume $2 π / 3$ ) and a cone (having volume $π / 3$ ).

use plots in
module()
local p1,p2,p3;
p1:=plot3d(2*cos(phi), theta=0..2*Pi,phi=0.. (1/4)*Pi,coords=spherical);
p2:=plot3d([rho, theta,Pi/4],rho=0.. sqrt(2),theta=0..2*Pi, coords = spherical); p3:=display(p1,p2, scaling=constrained, axes=frame,labels=[x,y,z],tickmarks=[3,3,3],orientation=[-30,80,0],lightmodel=none);
print(p3);
end module:
end use:

Figure 7.6.2(a) The region $R$

•	The upper limit of the innermost integral is obtained by using $z = \sqrt{x^{2} + y^{2}}$ to rewrite $x^{2} + y^{2} + {(z - 1)}^{2} = 1$ as $ρ^{2} - 2 z = ρ (ρ - 2 \cos (φ)) = 0$ and solving for $ρ = 2 \cos (φ)$ .

•	The upper limit of the middle integral is obtained by noting that for the cone $z = \sqrt{x^{2} + y^{2}}$ , $φ = π / 4$ .

Maple Solution - Interactive

Table 7.6.2(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 7.6.2(a) Solution by visualization task template

Table 7.6.2(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)

$ρ_{2} (φ, θ)$

>	$2 \cos (φ)$

$2 \cos (φ)$

(3)

$φ_{1} (θ)$

$0$

(4)

$φ_{2} (θ)$

$π / 4$

$\frac{1}{4} π$

(5)

$a$

$0$

(6)

$b$

$2 π$

(7)

Inert Integral: $d ρ d φ d θ$

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

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

(8)

Value:

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

$π$

(9)

Stepwise Evaluation:

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

$\begin{array}{c} \int_{0}^{2 π} \int_{0}^{\frac{π}{4}} \int_{0}^{2 \cos (φ)} ρ^{2} \sin (φ) &DifferentialD; ρ &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} \int_{0}^{\frac{π}{4}} (\binom{\frac{ρ^{3} \sin (φ)}{3}}{} | \binom{}{ρ = 0 .. 2 \cos (φ)}) &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} \int_{0}^{\frac{π}{4}} \frac{8 \sin (φ) {\cos (φ)}^{3}}{3} &DifferentialD; φ &DifferentialD; θ \\ = & \int_{0}^{2 π} (\binom{- \frac{2 {\cos (φ)}^{4}}{3}}{} | \binom{}{φ = 0 .. \frac{π}{4}}) &DifferentialD; θ \\ = & \int_{0}^{2 π} \frac{1}{2} &DifferentialD; θ \\ = & \binom{\frac{θ}{2}}{} | \binom{}{θ = 0 .. 2 π} \end{array}$

$π$

(10)

Table 7.6.2(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

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

Table 7.6.2(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

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

Table 7.6.2(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

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

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

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

$π$

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

Solution via the top-level Int and int commands

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

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

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