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Chapter 8: Applications of Triple Integration

Section 8.1: Volume

Example 8.1.8

Use an iterated triple integral to obtain the volume of $R$ , the region common to the two cylinders $x^{2} + y^{2} = 1$ and $x^{2} + z^{2} = 1$ .

Solution

Mathematical Solution

Figure 8.1.8(a) shows the two intersecting cylinders; Figure 8.1.8(b), the actual region $R$ ; and Figure 8.1.8(c), a cut-away view in the first octant.

>	use plottools, plots in EX818:=module() local p1,p2; export p3; p1:=cylinder([0,0,-2],1,4); p2:=rotate(p1,0,Pi/2,0); p3:=display(p1,p2,scaling=constrained,labels=[x,y,z],axes=frame,tickmarks=[5,2,5],lightmodel=light4,orientation=[-75,65,0]); print(p3); end module: end use:

Figure 8.1.8(a) Cylinders

Figure 8.1.8(b) Region $R$

Figure 8.1.8(c) Cut-away

The volume of $R$ can be found by iterating in the order $dz dy dx$ , with the horizontal cylinder $x^{2} + z^{2} = 1$ solved for $z = \pm \sqrt{1 - x^{2}}$ and the vertical cylinder $x^{2} + y^{2} = 1$ solved for $y = \pm \sqrt{1 - x^{2}}$ . As a result, the requisite volume is then

$\int_{- 1}^{1} \int_{- \sqrt{1 - x^{2}}}^{{\sqrt{1 - x^{2}}}_{}} \int_{- \sqrt{1 - x^{2}}}^{{\sqrt{1 - x^{2}}}_{}} 1 dz dy dx$ = $\frac{16}{3}$

Maple Solution - Interactive

Table 8.1.8(a) provides, via a visualization task template, a solution in Cartesian coordinates.

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 3-D

Evaluate $∭_{R} Ψ (x, y, z) dv$ and Graph $R$

Volume Element $dv$

, where $Ψ =$

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

Table 8.1.8(a) Solution in Cartesian coordinates via a visualization task template

Table 8.1.8(b) provides, via a visualization task template, a solution in cylindrical coordinates.

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical

Evaluate $∭_{R} Ψ (r, θ, z) dv$ and Graph $R$

Volume Element $dv$

$r dz dr d θ$

$r dz d θ dr$

$r dr d θ dz$

$r dr dz d θ$

$r d θ dr dz$

$r d θ dz dr$

, where $Ψ =$

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

Table 8.1.8(b) Solution in cylindrical coordinates via a visualization task template

Table 8.1.8(c) provides, via a task template that implements the MultiInt command from the Student MultivariateCalculus package, a solution in Cartesian coordinates.

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Multiple Integration≻Cartesian 3-D

Iterated Triple Integrals in Cartesian Coordinates

Integrand:

$1$

(1)

Region: $\{z_{1} (x, y) \leq z \leq z_{2} (x, y), y_{1} (x) \leq y \leq y_{2} (x), a \leq x \leq b\}$

$z_{1} (x, y)$

>	$- \sqrt{1 - x^{2}}$

$- \sqrt{- x^{2} + 1}$

(2)

$z_{2} (x, y)$

>	$\sqrt{1 - x^{2}}$

$\sqrt{- x^{2} + 1}$

(3)

$y_{1} (x)$

>	$- \sqrt{1 - x^{2}}$

$- \sqrt{- x^{2} + 1}$

(4)

$y_{2} (x)$

>	$\sqrt{1 - x^{2}}$

$\sqrt{- x^{2} + 1}$

(5)

$a$

$- 1$

(6)

$b$

$1$

(7)

Inert Integral: $dz dy dx$

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., y = .., x = .., output = integral)$

$\int_{- 1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x$

(8)

Value:

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., y = .., x = ..)$

$\frac{16}{3}$

(9)

Stepwise Evaluation:

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., y = .., x = .., output = steps)$

$\begin{array}{c} \int_{−1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x \\ = & \int_{−1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} (\binom{z}{} | \binom{}{z = - \sqrt{- x^{2} + 1} .. \sqrt{- x^{2} + 1}}) &DifferentialD; y &DifferentialD; x \\ = & \int_{−1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 2 \sqrt{- x^{2} + 1} &DifferentialD; y &DifferentialD; x \\ = & \int_{−1}^{1} (\binom{2 \sqrt{- x^{2} + 1} y}{} | \binom{}{y = - \sqrt{- x^{2} + 1} .. \sqrt{- x^{2} + 1}}) &DifferentialD; x \\ = & \int_{−1}^{1} (- 4 x^{2} + 4) &DifferentialD; x \\ = & \binom{(- \frac{4}{3} x^{3} + 4 x)}{} | \binom{}{x = −1 .. 1} \end{array}$

$\frac{16}{3}$

(10)

Table 8.1.8(c) Task template implementation of MultiInt solution in Cartesian coordinates

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$c = \sqrt{1 - x^{2}}$ $\overset{assign}{\to}$

•	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

$1$ $\overset{MultiInt}{\to}$ $\int_{- 1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x$ $=$ $\frac{16}{3}$

Table 8.1.8(d) provides, via a task template that implements the MultiInt command from the Student MultivariateCalculus package, a solution in cylindrical coordinates.

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

Iterated Triple Integral in Cylindrical Coordinates

Integrand:

$1$

(11)

Region: $\{z_{1} (r, θ) \leq z \leq z_{2} (r, θ), r_{1} (θ) \leq r \leq r_{2} (θ), a \leq θ \leq b\}$

$z_{1} (r, θ)$

>	$- \sqrt{1 - r^{2} \cos^{2} (θ)}$

$- \sqrt{1 - r^{2} {\cos (θ)}^{2}}$

(12)

$z_{2} (r, θ)$

>	$\sqrt{1 - r^{2} \cos^{2} (θ)}$

$\sqrt{1 - r^{2} {\cos (θ)}^{2}}$

(13)

$r_{1} (θ)$

$0$

(14)

$r_{2} (θ)$

$1$

(15)

$a$

$0$

(16)

$b$

$2 π$

(17)

Inert Integral: $dz dr d θ$

(Note automatic insertion of Jacobian.)

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., r = .., θ = .., coordinates = cylindrical [r, θ, z], output = integral)$

$\int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2} {\cos (θ)}^{2}}}^{\sqrt{1 - r^{2} {\cos (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ$

(18)

Value:

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., r = .., θ = .., coordinates = cylindrical [r, θ, z])$

$\frac{16}{3}$

(19)

Stepwise Evaluation:

>	$Student [MultivariateCalculus] [MultiInt] (, z = .., r = .., θ = .., coordinates = cylindrical [r, θ, z], output = steps)$

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

$\frac{16}{3}$

(20)

Table 8.1.8(d) Task template implementation of MultiInt solution in cylindrical coordinates

Access the MultiInt command via the Context Panel

$C = \sqrt{1 - r^{2} \cos^{2} (θ)}$ $\overset{assign}{\to}$

•	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

$1$ $\overset{MultiInt}{\to}$ $\int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2} {\cos (θ)}^{2}}}^{\sqrt{1 - r^{2} {\cos (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ$ $=$ $\frac{16}{3}$

Table 8.1.8(e) provides solutions from first principles: on the left, a solution in Cartesian coordinates; on the right, in cylindrical coordinates.

$\int_{- 1}^{1} \int_{- \sqrt{1 - x^{2}}}^{{\sqrt{1 - x^{2}}}_{}} \int_{- \sqrt{1 - x^{2}}}^{{\sqrt{1 - x^{2}}}_{}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x$ = $\frac{16}{3}$	$\int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2} \cos^{2} (θ)}}^{{\sqrt{1 - r^{2} \cos^{2} (θ)}}_{}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ$ = $\frac{16}{3}$
Table 8.1.8(e) From first principles, solutions in both Cartesian and cylindrical coordinates

Maple Solution - Coded

Table 8.1.8(f) provides, from first principles using the top-level Int and int commands, solutions in Cartesian and cylindrical coordinates.

$Int (1, [z = - \sqrt{1 - x^{2}} .. \sqrt{1 - x^{2}}, y = - \sqrt{1 - x^{2}} .. \sqrt{1 - x^{2}}, x = - 1 .. 1]) = int (1, [z = - \sqrt{1 - x^{2}} .. \sqrt{1 - x^{2}}, y = - \sqrt{1 - x^{2}} .. \sqrt{1 - x^{2}}, x = - 1 .. 1])$

$\int_{- 1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x = \frac{16}{3}$

$Int (r, [z = - \sqrt{1 - r^{2} \cos^{2} (θ)} .. \sqrt{1 - r^{2} \cos^{2} (θ)}, r = 0 .. 1, θ = 0 .. 2 π]) = int (r, [z = - \sqrt{1 - r^{2} \cos^{2} (θ)} .. \sqrt{1 - r^{2} \cos^{2} (θ)}, r = 0 .. 1, θ = 0 .. 2 π])$

$\int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2} {\cos (θ)}^{2}}}^{\sqrt{1 - r^{2} {\cos (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ = \frac{16}{3}$

Table 8.1.8(f) From first principles, solutions in Cartesian and cylindrical coordinates.

Table 8.1.8(g) demonstrates the syntax applying the MultiInt command in both Cartesian and cylindrical coordinates.

Initialize

•	Install the Student MultivariateCalculus package.

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

•	Context Panel: Assign Name

$g = \sqrt{1 - x^{2}}$ $\overset{assign}{\to}$

•	Context Panel: Assign Name

$h = \sqrt{1 - r^{2} \cos^{2} (θ)}$ $\overset{assign}{\to}$

Implement the MultiInt command in Cartesian coordinates

$MultiInt (1, z = - g .. g, y = - g .. g, x = - 1 .. 1, output = integral) = MultiInt (1, z = - g .. g, y = - g .. g, x = - 1 .. 1)$

$\int_{- 1}^{1} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} \int_{- \sqrt{- x^{2} + 1}}^{\sqrt{- x^{2} + 1}} 1 &DifferentialD; z &DifferentialD; y &DifferentialD; x = \frac{16}{3}$

Implement the MultiInt command in cylindrical coordinates

$MultiInt (1, z = - h .. h, r = 0 .. 1, θ = 0 .. 2 π, coordinates = cylindrical [r, θ, z], output = integral) = MultiInt (1, z = - h .. h, r = 0 .. 1, θ = 0 .. 2 π, coordinates = cylindrical [r, θ, z])$

$\int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2} {\cos (θ)}^{2}}}^{\sqrt{1 - r^{2} {\cos (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ = \frac{16}{3}$

Table 8.1.8(g) Application of the MultiInt command in Cartesian and cylindrical coordinates

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