Calculus III 14.06 Triple Integrals and Applications

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14.06 Triple Integrals and Applications

• Use a triple integral to find the volume for a solid region.
• Find the mass center and inertia moments for a solid region.

Triple Integrals

Figure 14.6.1

The rules for double integrals can be extended to triple integrals. Consider a function \(f\) with three variables that is continuous over a bounded solid region \(Q\). Enclose \(Q\) with equal sized boxes and form an inner partition with all the boxes that lay entirely within \(Q\), as shown in Figure 14.6.1. The volume for the \(i\)th box is \(\Delta V_{i} = \Delta x_{i} \Delta y_{i} \Delta z_{i}. \:\:\:\: \color{red}{ \text{Volume for the }i \text{th box}}\) The norm \(\| \Delta \|\) for the partition is the length for the longest diagonal for the \(n\) boxes in the partition. Choose a point \((x_{i},y_{i},z_{i})\) in each box and form the Riemann sum $$\sum_{i=1}^{n} f(x_{i},y_{i},z_{i}) \Delta V_{i}. $$ Taking the limit at \(\| \Delta \| \rightarrow 0\) produces Definition 14.6.1. Definition 14.6.1 Triple Integral If \(f\) is continuous over a bounded solid region \(Q\), then the triple integral for \(f\) over \(Q\) is defined as $$ \int \int_{Q} \int f(x,y,z) \: dV = \lim_{\| \Delta \| \rightarrow 0} \sum_{i=1}^{n} f(x_{i},y_{i},z_{i}) \Delta V_{i} $$ provided the limit exists. The volume for the solid region \(Q\) is given by $$ Volume \: for \: Q = \int \int_{Q} \int \: dV.$$ Theorem 14.2.1 properties can be restated for triple integrals. 1. $$ \int \int_{Q} \int c f(x,y,z) \: dV$$ $$= c\int \int_{Q} \int f(x,y,z) \: dV$$ 2. $$ \int \int_{Q} \int [f(x,y,z) \pm g(x,y,z)] \: dV$$ $$= \int \int_{Q} \int f(x,y,z) \: dV \pm\int \int_{Q} \int g(x,y,z) \: dV$$ 3. $$ \int \int_{Q} \int f(x,y,z) \: dV $$ $$= \int \int_{Q_{1}} \int f(x,y,z) \: dV + \int \int_{Q_{2}} \int f(x,y,z) \: dV$$ The region \(Q\) is the union between two nonoverlapping solid subregions \(Q_{1}\) and \(Q_{2}\). If the solid region \(Q\) is simple, then the triple integral \( \int \int_{Q} \int f(x,y,z) \: dV\) can be evaluated with an iterated integral by choosing one from six possible integration orders: \( dx \: dy \: dz \:\:\:\: \) \( dy \: dx \: dz \:\:\:\: \) \( dz \: dx \: dy \) \( dx \: dz \: dy \:\:\:\: \) \( dy \: dz \: dx \:\:\:\: \) \( dz \: dy \: dx. \)

Theorem 14.6.1 Fubini's Theorem[1] On \(dz \: dy \: dx\) Integration Order

Left \(f\) be continuous on a solid region \(Q\) defined by

\( a \leqslant x \leqslant b,\)

\(h_{1}(x) \leqslant y \leqslant h_{2}(x), \)

\(g_{1}(x,y) \leqslant z \leqslant g_{2}(x,y) \)

where \(h_{1}, \: h_{2}, \: g_{1}\) and \(g_{2}\) are continuous functions. This produces,

$$ \int \int_{Q} \int f(x,y,z) \: dV = \int_{a}^{b} \int_{h_{1}(x)}^{h_{2}(x)} \int_{g_{1}(x)}^{g_{2}(x)} f(x,y,z) \: dz \: dy \: dx. $$

To evaluate a triple iterated integral in the order \( dz \: dy \: dx\), hold both \(x\) and \(y\) constant for the innermost integration. Then, hold x constant for the second integration.

Example 14.6.1 Evaluating a Triple Iterated Integral

Evaluate the triple iterated integral

$$ \int_{0}^{2} \int_{0}^{x} \int_{0}^{x+y} e^{x}(y+2z) \: dz \: dy \: dx. $$

Solution For the first integration, hold \(x\) and \(y\) constant and integrate with respect to \(z\).

$$\int_{0}^{2} \int_{0}^{x} \int_{0}^{x+y} e^{x}(y+2z) \: dz \: dy \: dx $$	$$= \left. \int_{0}^{2} \int_{0}^{x} e^{x}(y+z^{2}) \right]_{0}^{x+y} \: dy \: dx $$
	$$= \int_{0}^{2} \int_{0}^{x} e^{x}(x^{2}+3xy+2y^{2}) \: dy \: dx $$

For the second integration, hold \(x\) constant and integrate with respect to \(y\).

$$ \int_{0}^{2} \int_{0}^{x} e^{x}(x^{2}+3xy+2y^{2}) \: dy \: dx$$	$$= \int_{0}^{2} \left[ e^{2} \left( x^{2}y+ \frac{3xy^{2}}{2} + \frac{2y^{3}}{3} \right) \right]_{0}^{x} \: dx $$
	$$= \frac{19}{6} \int_{0}^{2} x^{3} e^{x} \: dx$$

For the third integration, integrate with respect to \(x\).

$$ \frac{19}{6} \int_{0}^{2} x^{3} e^{x} \: dx $$	$$= \frac{19}{6} \left[ \vphantom{\frac{1}{2}} e^{x}(x^{3}-3x^{2}+6x-6) \right]_{0}^{2}$$
	$$= 19 \left( \frac{e^{x}}{3}+1 \right) \approx 65.797$$

Solid region \(Q\) lies between two surfaces. Figure 14.6.2

To find the limits for a particular integration order determine the innermost limits, which may be functions for the outer two variables. Then, by projecting the solid \(Q\) onto the coordinate plane for the outer two variables, determine their integration limits by the methods used for double integrals. For instance, to evaluate $$\int \int_{Q} \int f(x,y,z) \: dz \: dy \: dx $$ first determine the limits for \(z\); the integral then has the form $$\int \int \left[ \int_{g_{1}(x,y)}^{g_{2}(x,y)} f(x,y,z) \: dz \right] \: dy \: dx. $$ By projecting the solid \(Q\) onto the \(xy\)-plane the limits for \(x\) and \(y\) become apparent, as shown in Figure 14.6.2.

Example 14.6.2 Using a Triple Integral to Find Volume

Figure 14.6.3 Figure 14.6.4

Find the volume for the ellipsoid given by \(4x^{2}+4y^{2}+z^{2}=16.\) Solution Because \(x\), \(y\), and \(z\) are equally represented in the equation, any integration order can be used with equal difficulty. The integration order here will be \(dz, \: dy, \: dx\) will be chosen. The ellipsoid lies evenly about the origin, as shown in Figure 14.6.3. Integrate the first octant and multiply by eight to find the total volume. Determining the bounds for \(z\) yields \(0 \leqslant z \leqslant 2 \sqrt{4-x^{2}-y^{2}} \) The bounds for \(x\) and \(z\) \(0 \leqslant x \leqslant 2 \) and \( 0 \leqslant y \leqslant \sqrt{4-x^{2}}, \) as shown by examining the cross-section displayed in Figure 14.6.4. This produces the volume for the ellipsoid $$ V$$ $$= \int \int_{Q} \int \: dV$$ $$=8 \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \int_{0}^{2 \sqrt{4-x^{2}-y^{2}}} \: dz \: dy \: dx$$ $$= 8 \left. \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} z \right]_{0}^{2 \sqrt{4-x^{2}-y^{2}}} \: dy \: dx$$ $$= 16 \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \sqrt{(4-x^{2}) -y^{2}} \: dy \: dx$$ $$= 8 \int_{0}^{2} \left[ y \sqrt{4-x^{2}-y^{2}} + (4-x^{2}) \arcsin \left( \frac{y}{4-x^{2}} \right) \right]_{0}^{\sqrt{4-x^{2}}} \: dx $$ $$= 8 \int_{0}^{2} [0+(4-x^{2}) \arcsin(1)-0-0 ] \:dx$$ $$= 8 \int_{0}^{2} (4-x^{2}) \frac{\pi}{2} \:dx$$ $$= 4\pi \left[ 4x-\frac{x^{3}}{3}\right]_{0}^{2}= \frac{64 \pi}{3}$$

Example 14.6.3 Changing the Integration Order

Figure 14.6.5

Evaluate $$ \int_{0}^{\sqrt{\pi/2}} \int_{x}^{\sqrt{\pi/2}} \int_{1}^{3} \sin(y^{2}) \: dz \: dy \: dx. $$ Solution Note the second integration in the given order is $$ 2 \int \sin(y^{2}) \:dy$$ which is not an elementary function. To avoid this problem, change the integration order to \(dz \: dx \: dy\), so that \(y\) is the new outer variable. The bounds for the solid region \(Q\) are $$0 \leqslant x \leqslant \sqrt{\frac{\pi}{2}}$$ $$ x \leqslant y \leqslant \sqrt{\frac{\pi}{2}}$$ $$ 1 \leqslant z \leqslant 3$$ Projecting \(Q\) in the \(xy\)-plane yields the bounds $$0 \leqslant y \leqslant \sqrt{\frac{\pi}{2}} $$ and $$ 0 \leqslant x \leqslant y,$$ as shown in Figure 14.6.5. Evaluating the triple integral using the new order \(dz \: dx \: dy\) produces $$ \int_{0}^{\sqrt{\pi/2}} \int_{0}^{y} \int_{1}^{3} \sin(y^{2}) \: dz \: dx \: dy$$ $$= \left. \int_{0}^{\sqrt{\pi/2}} \int_{0}^{y} z \sin(y^{2}) \right]_{1}^{3} \: dx \: dy$$ $$= 2\int_{0}^{\sqrt{\pi/2}} \int_{0}^{y} \sin(y^{2}) \: dx \: dy $$ $$= 2 \left. \int_{0}^{\sqrt{\pi/2}} x \sin(y^{2}) \right]_{0}^{y} \: dy$$ $$= 2 \int_{0}^{\sqrt{\pi/2}} y \sin(y^{2}) \: dy$$ $$= \left. \vphantom{\frac{1}{2}} - \cos(y^{2}) \right]_{0}^{\sqrt{\pi/2}} = 1 $$

Example 14.6.4 Determining the Integration Limits

Figure 14.6.6 Figure 14.6.7 Figure 14.6.8

For solid regions a, b, and c, set up a triple integral for each volume. a. The region in the first octant bounded above by the cylinder \( z=1-y^{2}\) and lying between the vertical planes \(x+y=1\) and \(x+y=3\). b. The upper hemisphere \(z=\sqrt{1-x^{2}-y^{2}} \) c. The region bounded below by the paraboloid \(z=x^{2}+y^{2} \) and above by the sphere \(x^{2}+y^{2}+z^{2}=6\) Solution a. Note that the solid is bounded below by the \(xy\)-plane \(z=0\) and above by the cylinder \(z=1-y^{2}\), as shown in Figure 14.6.6. This yields the bounds for \(z\) \( 0 \leqslant z \leqslant 1-y^{2}. \:\:\:\: \color{red}{ \text{Bounds for }z}\) Projecting the region onto the \(xy\)-plane produces a parallelogram. Because the parallelogram has two sides that are parallel to the \(x\)-axis, the bounds for \(x\) and \(y\) are \(1-y \leqslant x \leqslant 3-y \) and \(0 \leqslant y \leqslant 1.\) The volume for the region is given by $$ V$$ $$= \int \int_{Q} \int \: dV $$ $$= \int_{0}^{1} \int_{1-y}^{3-y} \int_{0}^{1-y^{2}} \: dz \:dx \: dy. $$ b. For the upper hemisphere \(z=\sqrt{1-x^{2}-y^{2}} \), the bounds for \(z\) are \( 0 \leqslant z \leqslant \sqrt{1-x^{2}-y^{2}}. \:\:\:\: \color{red}{ \text{Bounds for }z}\) Note that the projection for the hemisphere onto the \(xy\)-plane is the circle \(x^{2}+y^{2}=1, \) as shown in Figure. 14.6.7. Either integration order, \(dx \: dy\) or \(dy \: dx\), is equally complex. Choosing the first produces \( -\sqrt{1-y} \leqslant x \leqslant \sqrt{1-y} \) and \(-1 \leqslant y \leqslant 1,\) bounds for \(x\) and \(y\). The volume for the region is given by $$ V$$ $$= \int \int_{Q} \int \: dV $$ $$= \int_{-1}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} \int_{0}^{\sqrt{1-x^{2}-y^{2}}} \: dz \:dx \: dy. $$ c. For the region bounded below by the paraboloid \(z=x^{2}+y^{2}\) and above by the sphere \(x^{2}+y^{2}+z^{2}=6\), the bounds for \(z\) are \( x^{2}+y^{2} \leqslant z \leqslant \sqrt{6 -x^{2} - y^{2} }. \:\:\:\: \color{red}{ \text{Bounds for }z}\) The sphere and paraboloid intersect at \(z=2\). The solid region \(Q\) projects onto the \(xy\)-plane onto the circle \( x^{2}+y^{2} = 2,\) as shown in Figure 14.6.8. Using the integration order \(dy \: dx\) produces the bounds \( -\sqrt{2-x^{2}} \leqslant y \leqslant \sqrt{2-x^{2}} \) and \( -\sqrt{2} \leqslant x \leqslant \sqrt{2},\) bounds for \(x\) and \(y\). The volume for the region is given by $$ V$$ $$= \int \int_{Q} \int \: dV $$ $$= \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^{2}}}^{\sqrt{2-x^{2}}} \int_{x^{2}+y^{2}}^{\sqrt{6-x^{2}-y^{2}}} \: dz \:dy \: dx. $$

Mass Center and Inertia Moments

Consider a solid region \(Q\) whose density is given by the density function \(\bf{\rho}\). The mass center for a solid region \(Q\) with mass \(m\) is given by \((\bar{x},\bar{y},\bar{z})\), where

$$m $$	$$= \int \int_{Q} \int \rho (x,y,z) \: dV $$	Mass for the solid
$$ M_{yz} $$	$$= \int \int_{Q} \int x \rho (x,y,z) \: dV $$	First moment about \(yz\)-plane
$$ M_{xz} $$	$$= \int \int_{Q} \int y \rho (x,y,z) \: dV $$	First moment about \(xz\)-plane
$$ M_{xy} $$	$$= \int \int_{Q} \int z \rho (x,y,z) \: dV $$	First moment about \(xy\)-plane
and
$$\bar{x}$$	$$= \frac{M_{yz}}{m}, \: \bar{y} = \frac{M_{xz}}{m}, \: \bar{z} = \frac{M_{xy}}{m}.$$

The quantities \( M_{yz} \), \( M_{xz} \), and \( M_{xy} \) are called the first moments for the region \(Q\) about the \(yz\)-, \(xz\)-, and \(xy\)-planes, respectively.

The first moments for solid regions are taken about a plane, whereas the second moments for solids are taken about a line. The second moments (or inertia moments) about the \(x\)-, \(y\)-, and \(z\)- axes are

$$I_{x} $$	$$ = \int \int_{Q} \int (y^{2}+z^{2}) \rho (x,y,z) \: dV \:\:\:\: $$	Inertia Moment about \(x\)-axis
$$I_{y} $$	$$= \int \int_{Q} \int (x^{2}+z^{2}) \rho (x,y,z) \: dV \:\:\:\: $$	Inertia Moment about \(y\)-axis
and
$$I_{z} $$	$$= \int \int_{Q} \int (x^{2}+y^{2}) \rho (x,y,z) \: dV. \:\:\:\: $$	Inertia Moment about \(z\)-axis

For problems requiring the calculation for all three moments, considerable effort can be saved by applying the additive property for triple integrals and writing

\(I_{x} = I_{xz} + I_{xy},\)

\(I_{y} = I_{yz} + I_{xy},\)

and

\(I_{z} = I_{yz} + I_{xz}\)

where \(I_{xy}, \: I_{xz}, \) and \(I_{yz}\) are

$$I_{xy} $$	$$ = \int \int_{Q} \int z^{2} \rho (x,y,z) \: dV $$
$$I_{xz} $$	$$= \int \int_{Q} \int y^{2} \rho (x,y,z) \: dV $$
and
$$I_{yz} $$	$$= \int \int_{Q} \int x^{2} \rho (x,y,z) \: dV. $$

Example 14.6.5 Finding the Mass Center for a Solid Region

Variable density: \(\rho(x,y,z)=k(x^{2}+y^{2}+z^{2})\) Figure 14.6.9

Find the mass center for the unit cube with a given density at the point \((x,y,z)\) is proportional to the distance squared from the origin, as shown in Figure 14.6.9. Solution Because the density at any point \((x,y,z)\) is proportional to the distance squared between \((0,0,0)\) and \((x,y,z)\), the density function is \( \rho (x,y,z) = k(x^{2}+y^{2}+z^{2}).\) Because the region is symmetric along all axes, any integration order will produce an integral with equal difficulty. $$m$$ $$= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} k (x^{2}+y^{2}+z^{2}) \: dz \: dy \: dx$$ $$=k \int_{0}^{1} \int_{0}^{1} \left[(x^{2}+y^{2})z+ \frac{z^{3}}{3} \right]_{0}^{1} \: dy \: dx $$ $$= k \int_{0}^{1} \int_{0}^{1} \left( x^{2}+y^{2}+\frac{1}{3} \right) \: dy \: dx $$ $$= k \int_{0}^{1} \left[ \left( x^{2}+\frac{1}{3}\right)y + \frac{y^{3}}{3} \right]_{0}^{1} \: dx $$ $$= k \int_{0}^{1} \left( x^{2}+\frac{2}{3}\right) \: dx$$ $$= k \left[ \frac{x^{3}}{3} + \frac{2x}{3} \right]_{0}^{1} = k$$ The first moment about the \(yz\)-plane is $$M_{yz}$$ $$= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} x (x^{2}+y^{2}+z^{2}) \: dz \: dy \: dx$$ $$=k \int_{0}^{1} x \left[ \int_{0}^{1} \int_{0}^{1} (x^{2}+y^{2}+z^{2}) \: dz \: dy \right] \: dx. $$ Note that \(x\) can be factored from the two inner integrals because it is a constant with respect to \(y\) and \(z\). After factoring, the two inner integrals are the same as for the mass \(m\). This produces the formula for the moment about the \(yz\)-axis. $$M_{yz} $$ $$= k \int_{0}^{1} x \left( x^{2}+\frac{2}{3} \right) \: dx$$ $$= k \left[ \frac{x^{4}}{4} + \frac{x^{2}}{3} \right]_{0}^{1} = \frac{7k}{12}.$$ This produces $$\bar{x}=\frac{M_{yz}}{m} = \frac{7k/12}{k}= \frac{7}{12} $$ Since the solid region is symmetric along all axes, then \(\bar{x} = \bar{y} = \bar{z}\), and the mass center is $$\left( \frac{7}{12},\frac{7}{12},\frac{7}{12}\right).$$

Example 14.6.6 Inertia Moments for a Solid Region

Variable density: \(\rho (x,y,z)=kz\) Figure 14.6.10

Find the inertia moments about the \(x\)- and \(y\)-axes for the solid region lying between the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the \(xy\)-plane. The density at any point \((x,y,z)\) is proportional to the distance between \((x,y,z)\) and the \(xy\)-plane. Solution The hemisphere is oriented over the origin with the \(z\)-axes vertical, as shown in Figure 14.6.10. The bounds are \(0 \leqslant z \leqslant \sqrt{4-x^{2}-y^{2}}\), \(-\sqrt{4-x^{2}} \leqslant y \leqslant \sqrt{4-x^{2}}\), and \(-2 \leqslant x \leqslant 2\). The density for the region is \(\rho (x,y,z)=kz,\) where \(k\) is the density. Since the solid is symmetric along the \(x\)- and \(y\)-axes, \(I_{x}=I_{y}\). $$ I_{x}$$ $$= \int \int_{Q} \int (y^{2}+z^{2}) \rho (x,y,z) \: dV $$ $$= \int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \int_{0}^{\sqrt{4-x^{2}-y^{2}}} (y^{2}+z^{2})(kz) \: dz \:dy \: dx$$ $$= k\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \left[ \frac{y^{2}z^{2}}{2} + \frac{z^{4}}{4} \right]_{0}^{\sqrt{4-x^{2}-y^{2}}} \:dy \: dx$$ $$= k\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \left[ \frac{y^{2}(4-x^{2}-y^{2})}{2} + \frac{(4-x^{2}-y^{2})^{2}}{4} \right] \:dy \: dx$$ $$= \frac{k}{4} \int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} [(4-x^{2})^{2}-y^{4} ] \:dy \: dx$$ $$= \frac{k}{4} \int_{-2}^{2} \left[ (4-x^{2})^{2}y - \frac{y^{5}}{5} \right]_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \: dx $$ $$= \frac{k}{4} \int_{-2}^{2} \frac{8}{5} (4-x^{2})^{5/2} \: dx $$ $$= \frac{4k}{5} \int_{0}^{2} (4-x^{2})^{5/2} \: dx$$     \(x = 2 \sin \theta\) $$= \frac{4k}{5} \int_{0}^{\pi/2} 64 \cos^{6} \theta \: d\theta$$ $$= \left( \frac{256k}{5}\right) \left( \frac{5\pi}{32} \right)$$     Wallis' Formula $$= 8k \pi.$$ Hence, \(I_{x}=8k \pi = I_{y}\). The moment about the \(z\)-axis is $$I_{z} = \frac{16}{3} k \pi. $$ This shows a greater resistance to rotation about the \(x\)- or \(y\)-axis than about the \(z\)-axis.

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Parent Article: Calculus III 14 Multiple Integration