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Chapter 8: Applications of Triple Integration

Section 8.1: Volume

Example 8.1.30

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

Solution

Mathematical Solution

Figure 8.1.30(a) shows the two intersecting cylinders; Figure 8.1.30(b), the actual region $R$ ; and Figure 8.1.30(c), one-quarter of the top-half of the region.

use plots, plottools in
EX8130:=module()
local p1,p2,p3,p4,p5;
export p6;
p1:=cylinder([0,0,-2],1,4,color=red);
p2:=cylinder([0,0,-2],1,4,color=blue);
p3:=cylinder([0,0,-2],1,4,color=green);
p4:=rotate(p2,0,(1/2)*Pi,0);
p5:=rotate(p3,(1/2)*Pi,0,0);
p6:=display(p1,p4,p5,scaling=constrained,labels=[x,y,z],axes=frame,tickmarks=[5,5,5],orientation=[-60,70,0]);
print(p6);
end module:
end use:

Figure 8.1.30(a) Cylinders

Figure 8.1.30(b) Region $R$

>	use plots in module() local p1,p2,p3; p1:=display(EX8130:-p6,view=[0..1,0..1,0..1],axes=box,orientation=[-75,65,0],tickmarks=[2,2,2],axes=frame): p2:=plot3d([x,x,z],x=0..1/sqrt(2),z=0..sqrt(1-x^2)): p3:=display(p1,p2,orientation=[-150,80,0],lightmodel=none); print(p3); end module: end use:

Figure 8.1.30(c) Cut-away

Although the volume of $R$ can be found by integrating in Cartesian coordinates, it turns out to be easier to work in cylindrical coordinates, iterating in the order $dz dr d θ$ . The complete volume of $R$ is then

$8 \int_{5 π / 4}^{7 π / 4} \int_{0}^{1} \int_{0}^{{\sqrt{1 - r^{2} \sin^{2} (θ)}}_{}} r dz dr d θ$ = $8 (2 - \sqrt{2})$ ≐ 4.69

Note the consistent coloring through Figures 8.1.30(1 - c). The cylinder $y^{2} + z^{2} = 1$ , parallel to the $x$ -axis, is drawn in blue in Figure 8.1.30(a), and parts of this cylinder appearing in the other two figures are likewise in blue. Similarly with the cylinder $x^{2} + z^{2} = 1$ , parallel to the $y$ -axis, and drawn in green; and with the cylinder $x^{2} + y^{2} = 1$ , parallel to the $z$ -axis, and drawn in red.

The upper bound on the inner integral is $z = \sqrt{1 - y^{2}}$ expressed in cylindrical coordinates. In Figure 8.1.30(c), the highest point on the blue portion of the surface is directly above the origin. The bounds on $θ$ are determined from the intersection of the blue and green cylinders by solving the equations $x^{2} + z^{2} = 1$ and $y^{2} + z^{2} - 1$ for $y = \pm x$ .

Maple Solution - Interactive

Table 8.1.30(a) provides, via a visualization task template, a solution in cylindrical coordinates. The volume found is one-eighth the total volume.

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 8.1.30(a) Solution in cylindrical coordinates via a visualization task template

Table 8.1.30(b) 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:

$1$

(1)

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

$z_{1} (r, θ)$

$0$

(2)

$z_{2} (r, θ)$

>	$\sqrt{1 - r^{2} \sin^{2} (θ)}$

$\sqrt{1 - r^{2} {\sin (θ)}^{2}}$

(3)

$r_{1} (θ)$

$0$

(4)

$r_{2} (θ)$

$1$

(5)

$a$

$5 π / 4$

$\frac{5 π}{4}$

(6)

$b$

$7 π / 4$

$\frac{7 π}{4}$

(7)

Inert Integral: $dz dr d θ$

(Note automatic insertion of Jacobian.)

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

$\int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} \int_{0}^{\sqrt{1 - r^{2} {\sin (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ$

(8)

Value:

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

$2 - \sqrt{2}$

(9)

Stepwise Evaluation:

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

$\begin{array}{c} \int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} \int_{0}^{\sqrt{1 - r^{2} {\sin (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ \\ = & \int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} (\binom{r z}{} | \binom{}{z = 0 .. \sqrt{1 - r^{2} {\sin (θ)}^{2}}}) &DifferentialD; r &DifferentialD; θ \\ = & \int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} r \sqrt{1 - r^{2} {\sin (θ)}^{2}} &DifferentialD; r &DifferentialD; θ \\ = & \int_{\frac{5 π}{4}}^{\frac{7 π}{4}} (\binom{- \frac{{(1 - r^{2} {\sin (θ)}^{2})}^{\frac{3}{2}}}{3 {\sin (θ)}^{2}}}{} | \binom{}{r = 0 .. 1}) &DifferentialD; θ \\ = & \int_{\frac{5 π}{4}}^{\frac{7 π}{4}} - \frac{{(1 - {\sin (θ)}^{2})}^{\frac{3}{2}} - 1}{3 {\sin (θ)}^{2}} &DifferentialD; θ \\ = & \binom{(- \frac{\cot (θ)}{3} + \frac{\cos (θ) ({\sin (θ)}^{2} + 1)}{3 \sin (θ) \sqrt{{\cos (θ)}^{2}}})}{} | \binom{}{θ = \frac{5 π}{4} .. \frac{7 π}{4}} \end{array}$

$2 - \sqrt{2}$

(10)

Table 8.1.30(b) Task template implementation of MultiInt solution in cylindrical coordinates

Table 8.1.30(c) provides a cylindrical-coordinate solution from first principles.

•	Calculus palette: Iterated triple integral template Complete the template as shown to the right.

•	Context Panel: Evaluate and Display Inline

$8 \int_{5 π / 4}^{7 π / 4} \int_{0}^{1} \int_{0}^{{\sqrt{1 - r^{2} \sin^{2} (θ)}}_{}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ$ = $16 - 8 \sqrt{2}$

Table 8.1.30(c) From first principles, a solution in cylindrical coordinates

Maple Solution - Coded

Table 8.1.30(d) provides, from first principles using the top-level Int and int commands, a solution in cylindrical coordinates.

$8 Int (r, [z = 0 .. \sqrt{1 - r^{2} \sin^{2} (θ)}, r = 0 .. 1, θ = 5 π / 4 .. 7 π / 4]) = 8 int (r, [z = 0 .. \sqrt{1 - r^{2} \sin^{2} (θ)}, r = 0 .. 1, θ = 5 π / 4 .. 7 π / 4])$

$8 (\int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} \int_{0}^{\sqrt{1 - r^{2} {\sin (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ) = 16 - 8 \sqrt{2}$

Table 8.1.30(d) From first principles, solutions in Cartesian and cylindrical coordinates.

Table 8.1.30(e) demonstrates the syntax applying the MultiInt command in cylindrical coordinates.

Initialize

•	Install the Student MultivariateCalculus package.

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

•	Context Panel: Assign Name

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

Implement the MultiInt command in cylindrical coordinates

$8 MultiInt (1, z = 0 .. h, r = 0 .. 1, θ = \frac{5 π}{4} .. \frac{7 π}{4}, coordinates = cylindrical [r, θ, z], output = integral)$

$8 (\int_{\frac{5 π}{4}}^{\frac{7 π}{4}} \int_{0}^{1} \int_{0}^{\sqrt{1 - r^{2} {\sin (θ)}^{2}}} r &DifferentialD; z &DifferentialD; r &DifferentialD; θ)$

$8 MultiInt (1, z = 0 .. h, r = 0 .. 1, θ = \frac{5 π}{4} .. \frac{7 π}{4}, coordinates = cylindrical [r, θ, z])$ = $16 - 8 \sqrt{2}$

Table 8.1.30(e) Application of the MultiInt command in cylindrical coordinates

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