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The ultimate virtual reality learning experience that improves student outcomes and increase student engagement through fearless learning. Minkowski space: each point has coords (x, y, z, ct); the metric is ds 2 = c 2 dt 2 – dx 2 – dy 2 – dz 2 dr 2 is invariant under rotation of coords.
The graph of all points $(x,y,f(x,y))$ with $(x,y)$ in this domain is an elliptic paraboloid, as shown in the following figure. Applet loading Graph of elliptic paraboloid.
We learned in Section 13.2 how to compute the signed volume (V) under a surface (z=f(x,y)) over a region (R): (V = iint_R f(x,y) dA). It follows naturally that if (f(x,y)geq g(x,y)) on (R), then the volume between (f(x,y)) and (g(x,y)) on (R) is
[ begin{align} V &= iint_R f(x,y) dA - iint_R g(x,y) dA [4pt] &= iint_R big(f(x,y)-g(x,y)big) dA. end{align} ]
theorem 124: Volume Between Surfaces
Let (f) and (g) be continuous functions on a closed, bounded region (R), where (f(x,y)geq g(x,y)) for all ((x,y)) in (R). The volume (V) between (f) and (g) over (R) is
[V =iint_R big(f(x,y)-g(x,y)big) dA.]
Example (PageIndex{1}): Finding volume between surfaces
Find the volume of the space region bounded by the planes (z=3x+y-4) and (z=8-3x-2y) in the (1^text{st}) octant. In Figure 13.36(a) the planes are drawn; in (b), only the defined region is given.
Solution
We need to determine the region (R) over which we will integrate. To do so, we need to determine where the planes intersect. They have common (z)-values when (3x+y-4=8-3x-2y). Applying a little algebra, we have:
[begin{align*}
3x+y-4 &= 8-3x-2y
6x+3y &=12
2x+y &=4
end{align*}]
The planes intersect along the line (2x+y=4). Therefore the region (R) is bounded by (x=0), (y=0), and (y=4-2x); we can convert these bounds to integration bounds of (0leq xleq 2), (0leq yleq 4-2x). Thus
[begin{align*}
V &= iint_R big(8-3x-2y-(3x+y-4)big) dA
&= int_0^2int_0^{4-2x} big(12-6x-3ybig) dy , dx
&= 16text{u}^3.
end{align*}]
The volume between the surfaces is (16) cubic units.
In the preceding example, we found the volume by evaluating the integral [ int_0^2int_0^{4-2x} big(8-3x-2y-(3x+y-4)big) dy , dx.] Note how we can rewrite the integrand as an integral, much as we did in Section 13.1:
[8-3x-2y-(3x+y-4) = int_{3x+y-4}^{8-3x-2y} dz.]
Thus we can rewrite the double integral that finds volume as
[int_0^2int_0^{4-2x} big(8-3x-2y-(3x+y-4)big) dy , dx = int_0^2int_0^{4-2x}left(int_{3x+y-4}^{8-3x-2y} dzright) dy , dx.]
This no longer looks like a ’double integral,’ but more like a ’triple integral.’ Just as our first introduction to double integrals was in the context of finding the area of a plane region, our introduction into triple integrals will be in the context of finding the volume of a space region.
To formally find the volume of a closed, bounded region (D) in space, such as the one shown in Figure 13.37(a), we start with an approximation. Break (D) into (n) rectangular solids; the solids near the boundary of (D) may possibly not include portions of (D) and/or include extra space. In Figure 13.37(b), we zoom in on a portion of the boundary of (D) to show a rectangular solid that contains space not in (D); as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids.
The volume (Delta V_i) of the (i^text{th}) solid (D_i) is (Delta V_i = Delta x_iDelta y_iDelta z_i), where (Delta x_i), (Delta y_i) and (Delta z_i) give the dimensions of the rectangular solid in the (x), (y) and (z) directions, respectively. By summing up the volumes of all (n) solids, we get an approximation of the volume (V) of (D):
$$V approx sum_{i=1}^n Delta V_i = sum_{i=1}^n Delta x_iDelta y_iDelta z_i.]
Let (||Delta D||) represent the length of the longest diagonal of rectangular solids in the subdivision of (D). As (||Delta D||to 0), the volume of each solid goes to 0, as do each of (Delta x_i), (Delta y_i) and (Delta z_i), for all (i). Our calculus experience tells us that taking a limit as (||Delta D||to 0) turns our approximation of (V) into an exact calculation of (V). Before we state this result in a theorem, we use a definition to define some terms.
Definition 106: Triple Integrals, Iterated Integration (Part I)
Let (D) be a closed, bounded region in space. Let (a) and (b) be real numbers, let (g_1(x)) and (g_2(x)) be continuous functions of (x), and let (f_1(x,y)) and (f_2(x,y)) be continuous functions of (x) and (y).
*The volume (V) of (D) is denoted by a triple integral, $$V = iiint_D dV.$$
*The iterated integral ( int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx) is evaluated as
$$int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx=int_a^bint_{g_1(x)}^{g_2(x)}left(int_{f_1(x,y)}^{f_2(x,y)} dzright) dy , dx.$$
Evaluating the above iterated integral is triple integration.
Our informal understanding of the notation (iiint_D dV) is ’sum up lots of little volumes over (D),’ analogous to our understanding of (iint_R dA) and (iint_R dm).
We now state the major theorem of this section.
theorem 125 Triple Integration (Part I)
Let (D) be a closed, bounded region in space and let (Delta D) be any subdivision of (D) into (n) rectangular solids, where the (i^text{th}) subregion (D_i) has dimensions (Delta x_itimesDelta y_itimesDelta z_i) and volume (Delta V_i).
*The volume (V) of (D) is
$$V = iiint_D dV = lim_{||Delta D||to0} sum_{i=1}^n Delta V_i = lim_{||Delta D||to0} sum_{i=1}^n Delta x_iDelta y_iDelta z_i.$$
*If (D) is defined as the region bounded by the planes (x=a) and (x=b), the cylinders (y=g_(x)) and (y=g_2(x)), and the surfaces (z=f_1(x,y)) and (z=f_2(x,y)), where (a<b), (g_1(x)leq g_2(x)) and (f_1(x,y)leq f_2(x,y)) on (D), then
$$iiint_D dV = int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx.$$
*(V) can be determined using iterated integration with other orders of integration (there are 6 total), as long as (D) is defined by the region enclosed by a pair of planes, a pair of cylinders, and a pair of surfaces.
We evaluated the area of a plane region (R) by iterated integration, where the bounds were ’from curve to curve, then from point to point.’ Theorem 125 allows us to find the volume of a space region with an iterated integral with bounds ’from surface to surface, then from curve to curve, then from point to point.’ In the iterated integral
[int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx,]
the bounds (aleq xleq b) and (g_1(x)leq yleq g_2(x)) define a region (R) in the (x)-(y) plane over which the region (D) exists in space. However, these bounds are also defining surfaces in space; (x=a) is a plane and (y=g_1(x)) is a cylinder. The combination of these 6 surfaces enclose, and define, (D).
Examples will help us understand triple integration, including integrating with various orders of integration.
Example (PageIndex{2}): Finding the volume of a space region with triple integration
Find the volume of the space region in the (1^{,st}) octant bounded by the plane (z=2-y/3-2x/3), shown in Figure 13.38(a), using the order of integration (dz , dy , dx). Set up the triple integrals that give the volume in the other 5 orders of integration.
Solution
Starting with the order of integration (dz , dy , dx), we need to first find bounds on (z). The region (D) is bounded below by the plane (z=0) (because we are restricted to the first octant) and above by (z=2-y/3-2x/3); (0leq zleq 2-y/3-2x/3).
To find the bounds on (y) and (x), we ’collapse’ the region onto the (x)-(y) plane, giving the triangle shown in Figure 13.38(b). (We know the equation of the line (y=6-2x) in two ways. First, by setting (z=0), we have (0 = 2-y/3-2x/3 Rightarrow y=6-2x). Secondly, we know this is going to be a straight line between the points ((3,0)) and ((0,6)) in the (x)-(y) plane.)
We define that region (R), in the integration order of (dy , dx), with bounds (0leq yleq 6-2x) and (0leq xleq 3). Thus the volume (V) of the region (D) is:
[begin{align*}
V &= iiint_D dV
&= int_0^3int_0^{6-2x}int_0^{2-frac 13y-frac 23x} dz dy dz
&= int_0^3int_0^{6-2x}left(int_0^{2-frac 13y-frac 23x} dzright) dy dz
&=int_0^3int_0^{6-2x}zBig|_0^{2-frac 13y-frac 23x} dy dz
&= int_0^3int_0^{6-2x}left(2-frac 13y-frac 23xright) dy dz. end{align*}]
From this step on, we are evaluating a double integral as done many times before. We skip these steps and give the final volume,
[= 6text{u}^3. ]
The order (dz , dx , dy):
Now consider the volume using the order of integration (dz , dx , dy). The bounds on (z) are the same as before, (0leq zleq 2-y/3-2x/3). Collapsing the space region on the (x)-(y) plane as shown in Figure 13.38(b), we now describe this triangle with the order of integration (dx dy). This gives bounds (0leq xleq 3-y/2) and (0leq yleq 6). Thus the volume is given by the triple integral
[V = int_0^6int_0^{3-frac12y}int_0^{2-frac13y-frac23x} dz , dx , dy.]
The order (dx , dy , dz):
Following our ’surface to surface(ldots)’ strategy, we need to determine the (x)-surfaces that bound our space region. To do so, approach the region ’from behind,’ in the direction of increasing (x). The first surface we hit as we enter the region is the (y)-(z) plane, defined by (x=0). We come out of the region at the plane (z=2-y/3-2x/3); solving for (x), we have (x= 3-y/2-3z/2). Thus the bounds on (x) are: (0leq xleq 3-y/2-3z/2).
Now collapse the space region onto the (y)-(z) plane, as shown in Figure 13.39(a). (Again, we find the equation of the line (z=2-y/3) by setting (x=0) in the equation (x=3-y/2-3z/2).) We need to find bounds on this region with the order (dy dz). The curves that bound (y) are (y=0) and (y=6-3z); the points that bound (z) are 0 and 2. Thus the triple integral giving volume is:
[begin{array}{cc}
begin{array}{c}
0leq xleq 3-y/2-3z/2
0leq yleq 6-3z
0leq zleq 2
end{array}
&
Rightarrow quad int_0^2int_0^{6-3z}int_0^{3-y/2-3z/2} dx , dy , dz.
end{array}
]
The order (dx , dz , dy):
The (x)-bounds are the same as the order above. We now consider the triangle in Figure 13.39(a) and describe it with the order (dz dy): (0leq zleq 2-y/3) and (0leq yleq 6). Thus the volume is given by:
$$begin{array}{cc}
begin{array}{c}
0leq xleq 3-y/2-3z/2
0leq zleq 2-y/3
0leq yleq 6
end{array}
&
Rightarrow quad int_0^6int_0^{2-y/3}int_0^{3-y/2-3z/2} dx , dz , dy.
end{array}
]
The order (dy dz dx):
We now need to determine the (y)-surfaces that determine our region. Approaching the space region from ’behind’ and moving in the direction of increasing (y), we first enter the region at (y=0), and exit along the plane (z= 2-y/3-2x/3). Solving for (y), this plane has equation (y = 6-2x-3z). Thus (y) has bounds (0leq yleq 6-2x-3z).
Now collapse the region onto the (x)-(z) plane, as shown in Figure 13.39(b). The curves bounding this triangle are (z=0) and (z=2-2x/3); (x) is bounded by the points (x=0) to (x=3). Thus the triple integral giving volume is:
$$begin{array}{cc}
begin{array}{c}
0leq yleq 6-2x-3z
0leq zleq 2-2x/3
0leq xleq 3
end{array}
&
Rightarrow quad int_0^3int_0^{2-2x/3}int_0^{6-2x-3z} dy dz dx.
end{array}
]
The order (dy , dx , dz):
The (y)-bounds are the same as in the order above. We now determine the bounds of the triangle in Figure 13.39(b) using the order (dy , dx , dz). (x) is bounded by (x=0) and (x=3-3z/2); (z) is bounded between (z=0) and (z=2). This leads to the triple integral:
[begin{array}{cc}
begin{array}{c}
0leq yleq 6-2x-3z
0leq xleq 3-3z/2
0leq zleq 2
end{array}
&
Rightarrow quad int_0^2int_0^{3-3z/2}int_0^{6-2x-3z} dy , dx , dz.
end{array}
]
This problem was long, but hopefully useful, demonstrating how to determine bounds with every order of integration to describe the region (D). In practice, we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best.
In the previous example, we collapsed the surface into the (x)-(y), (x)-(z), and (y)-(z) planes as we determined the ’curve to curve, point to point’ bounds of integration. Since the surface was a triangular portion of a plane, this collapsing, or projecting, was simple: the projection of a straight line in space onto a coordinate plane is a line.
The following example shows us how to do this when dealing with more complicated surfaces and curves.
Example (PageIndex{3}): Finding the projection of a curve in space onto the coordinate planes
Consider the surfaces (z=3-x^2-y^2) and (z=2y), as shown in Figure 13.40(a). The curve of their intersection is shown, along with the projection of this curve into the coordinate planes, shown dashed. Find the equations of the projections into the coordinate planes.
Solution
The two surfaces are (z=3-x^2-y^2) and (z=2y). To find where they intersect, it is natural to set them equal to each other: (3-x^2-y^2=2y). This is an implicit function of (x) and (y) that gives all points ((x,y)) in the (x)-(y) plane where the (z) values of the two surfaces are equal.
We can rewrite this implicit function by completing the square:
$$3-x^2-y^2=2y quad Rightarrow quad y^2+2y+x^2=3quad Rightarrow quad (y+1)^2+x^2=4.$$
Thus in the (x)-(y) plane the projection of the intersection is a circle with radius 2, centered at ((0,-1)).
To project onto the (x)-(z) plane, we do a similar procedure: find the (x) and (z) values where the (y) values on the surface are the same. We start by solving the equation of each surface for (y). In this particular case, it works well to actually solve for (y^2):
(z=3-x^2-y^2 quad Rightarrow quad y^2=3-x^2-z)
(z=2y quad Rightarrow quad y^2=z^2/4).
Thus we have (after again completing the square):
$$3-x^2-z = z^2/4 quad Rightarrowquad frac{(z+2)^2}{16}+frac{x^2}4=1,$$
and ellipse centered at ((0,-2)) in the (x)-(z) plane with a major axis of length 8 and a minor axis of length 4.
Finally, to project the curve of intersection into the (y)-(z) plane, we solve equation for (x). Since (z=2y) is a cylinder that lacks the variable (x), it becomes our equation of the projection in the (y)-(z) plane.
All three projections are shown in Figure 13.40(b).
Example (PageIndex{4}): Finding the volume of a space region with triple integration
Set up the triple integrals that find the volume of the space region (D) bounded by the surfaces (x^2+y^2=1), (z=0) and (z=-y), as shown in Figure 13.41(a), with the orders of integration (dz , dy , dx), (dy , dx , dz) and (dx , dz , dy).
Solution
The order (dz , dy , dx):
The region (D) is bounded below by the plane (z=0) and above by the plane (z=-y). The cylinder (x^2+y^2=1) does not offer any bounds in the (z)-direction, as that surface is parallel to the (z)-axis. Thus (0leq zleq -y).
Collapsing the region into the (x)-(y) plane, we get part of the circle with equation (x^2+y^2=1) as shown in Figure 13.41(b). As a function of (x), this half circle has equation (y=-sqrt{1-x^2}). Thus (y) is bounded below by (-sqrt{1-x^2}) and above by (y=0): (-sqrt{1-x^2}leq yleq 0). The (x) bounds of the half circle are (-1leq xleq 1). All together, the bounds of integration and triple integral are as follows:
[begin{array}{cc}
begin{array}{c}
0leq zleq -y
-sqrt{1-x^2}leq yleq 0
-1leq xleq 1
end{array}
&
Rightarrow quad int_{-1}^1int_{-sqrt{1-x^2}}^{0}int_0^{-y} dz , dy , dx.
end{array}
]
We evaluate this triple integral:
[begin{align*}
int_{-1}^1int_{-sqrt{1-x^2}}^{0}int_0^{-y} dz , dy , dx &= int_{-1}^1int_{-sqrt{1-x^2}}^{0}big(-ybig) dy , dx
&=int_{-1}^1big(-frac12y^2big)Big|_{-sqrt{1-x^2}}^{0} dx
&= int_{-1}^1 frac12big(1-x^2big) dx
&= left.left(frac12left(x-frac13x^3right)right)right|_{-1}^1
&= frac23text{units}^3.
end{align*}]
With the order (dy , dx , dz):
The region is bounded ’below’ in the (y)-direction by the surface (x^2+y^2=1 Rightarrow y=-sqrt{1-x^2}) and ’above’ by the surface (y=-z). Thus the (y) bounds are (-sqrt{1-x^2}leq yleq -z).
Collapsing the region onto the (x)-(z) plane gives the region shown in Figure 13.42(a); this half circle has equation (x^2+z^2=1). (We find this curve by solving each surface for (y^2), then setting them equal to each other. We have (y^2=1-x^2) and (y=-zRightarrow y^2=z^2). Font free download. Thus (x^2+z^2=1).) It is bounded below by (x=-sqrt{1-z^2}) and above by (x=sqrt{1-z^2}), where (z) is bounded by (0leq zleq 1). All together, we have:
[begin{array}{cc}
begin{array}{c}
-sqrt{1-x^2}leq yleq -z
-sqrt{1-z^2}leq xleq sqrt{1-z^2}
0leq zleq 1
end{array}
&
Rightarrow quad int_{0}^1int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{-sqrt{1-x^2}}^{-z} dy , dx , dz.
end{array}
]
With the order (dx , dz , dy):
(D) is bounded below by the surface (x=-sqrt{1-y^2}) and above by (sqrt{1-y^2}). We then collapse the region onto the (y)-(z) plane and get the triangle shown in Figure 13.42}(b). (The hypotenuse is the line (z=-y), just as the plane.) Thus (z) is bounded by (0leq zleq -y) and (y) is bounded by (-1leq yleq 0). This gives:
[begin{array}{cc}
begin{array}{c}
-sqrt{1-y^2}leq xleq sqrt{1-y^2}
0leq zleq -y
-1leq yleq 0
end{array}
&
Rightarrow quad int_{-1}^0int_{0}^{-y}int_{-sqrt{1-y^2}}^{sqrt{1-y^2}} dx , dz , dy.
end{array}
]
The following theorem states two things that should make ’common sense’ to us. First, using the triple integral to find volume of a region (D) should always return a positive number; we are computing volume here, not signed volume. Secondly, to compute the volume of a ’complicated’ region, we could break it up into subregions and compute the volumes of each subregion separately, summing them later to find the total volume.
THEOREM 126: Properties of Triple Integrals
Let (D) be a closed, bounded region in space, and let (D_1) and (D_2) be non-overlapping regions such that (D=D_1bigcup D_2).
*( iiint_D dV geq 0)
*( iiint_D dV = iiint_{D_1} dV + iiint_{D_2} dV.)
We use this latter property in the next example.
Example (PageIndex{5}): Finding the volume of a space region with triple integration
Find the volume of the space region (D) bounded by the coordinate planes, (z=1-x/2) and (z=1-y/4), as shown in Figure 13.43(a). Set up the triple integrals that find the volume of (D) in all 6 orders of integration.
Solution
Following the bounds--determining strategy of ’surface to surface, curve to curve, and point to point,’ we can see that the most difficult orders of integration are the two in which we integrate with respect to (z) first, for there are two ’upper’ surfaces that bound (D) in the (z)-direction. So we start by noting that we have
[0leq zleq 1-frac12x quadtext{and}quad 0leq zleq 1-frac14y.]
We now collapse the region (D) onto the (x)-(y) axis, as shown in Figure 13.43(b). The boundary of (D), the line from ((0,0,1)) to ((2,4,0)), is shown in part (b) of the figure as a dashed line;
https://diarynote-jp.indered.space
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The ultimate virtual reality learning experience that improves student outcomes and increase student engagement through fearless learning. Minkowski space: each point has coords (x, y, z, ct); the metric is ds 2 = c 2 dt 2 – dx 2 – dy 2 – dz 2 dr 2 is invariant under rotation of coords.
The graph of all points $(x,y,f(x,y))$ with $(x,y)$ in this domain is an elliptic paraboloid, as shown in the following figure. Applet loading Graph of elliptic paraboloid.
We learned in Section 13.2 how to compute the signed volume (V) under a surface (z=f(x,y)) over a region (R): (V = iint_R f(x,y) dA). It follows naturally that if (f(x,y)geq g(x,y)) on (R), then the volume between (f(x,y)) and (g(x,y)) on (R) is
[ begin{align} V &= iint_R f(x,y) dA - iint_R g(x,y) dA [4pt] &= iint_R big(f(x,y)-g(x,y)big) dA. end{align} ]
theorem 124: Volume Between Surfaces
Let (f) and (g) be continuous functions on a closed, bounded region (R), where (f(x,y)geq g(x,y)) for all ((x,y)) in (R). The volume (V) between (f) and (g) over (R) is
[V =iint_R big(f(x,y)-g(x,y)big) dA.]
Example (PageIndex{1}): Finding volume between surfaces
Find the volume of the space region bounded by the planes (z=3x+y-4) and (z=8-3x-2y) in the (1^text{st}) octant. In Figure 13.36(a) the planes are drawn; in (b), only the defined region is given.
Solution
We need to determine the region (R) over which we will integrate. To do so, we need to determine where the planes intersect. They have common (z)-values when (3x+y-4=8-3x-2y). Applying a little algebra, we have:
[begin{align*}
3x+y-4 &= 8-3x-2y
6x+3y &=12
2x+y &=4
end{align*}]
The planes intersect along the line (2x+y=4). Therefore the region (R) is bounded by (x=0), (y=0), and (y=4-2x); we can convert these bounds to integration bounds of (0leq xleq 2), (0leq yleq 4-2x). Thus
[begin{align*}
V &= iint_R big(8-3x-2y-(3x+y-4)big) dA
&= int_0^2int_0^{4-2x} big(12-6x-3ybig) dy , dx
&= 16text{u}^3.
end{align*}]
The volume between the surfaces is (16) cubic units.
In the preceding example, we found the volume by evaluating the integral [ int_0^2int_0^{4-2x} big(8-3x-2y-(3x+y-4)big) dy , dx.] Note how we can rewrite the integrand as an integral, much as we did in Section 13.1:
[8-3x-2y-(3x+y-4) = int_{3x+y-4}^{8-3x-2y} dz.]
Thus we can rewrite the double integral that finds volume as
[int_0^2int_0^{4-2x} big(8-3x-2y-(3x+y-4)big) dy , dx = int_0^2int_0^{4-2x}left(int_{3x+y-4}^{8-3x-2y} dzright) dy , dx.]
This no longer looks like a ’double integral,’ but more like a ’triple integral.’ Just as our first introduction to double integrals was in the context of finding the area of a plane region, our introduction into triple integrals will be in the context of finding the volume of a space region.
To formally find the volume of a closed, bounded region (D) in space, such as the one shown in Figure 13.37(a), we start with an approximation. Break (D) into (n) rectangular solids; the solids near the boundary of (D) may possibly not include portions of (D) and/or include extra space. In Figure 13.37(b), we zoom in on a portion of the boundary of (D) to show a rectangular solid that contains space not in (D); as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids.
The volume (Delta V_i) of the (i^text{th}) solid (D_i) is (Delta V_i = Delta x_iDelta y_iDelta z_i), where (Delta x_i), (Delta y_i) and (Delta z_i) give the dimensions of the rectangular solid in the (x), (y) and (z) directions, respectively. By summing up the volumes of all (n) solids, we get an approximation of the volume (V) of (D):
$$V approx sum_{i=1}^n Delta V_i = sum_{i=1}^n Delta x_iDelta y_iDelta z_i.]
Let (||Delta D||) represent the length of the longest diagonal of rectangular solids in the subdivision of (D). As (||Delta D||to 0), the volume of each solid goes to 0, as do each of (Delta x_i), (Delta y_i) and (Delta z_i), for all (i). Our calculus experience tells us that taking a limit as (||Delta D||to 0) turns our approximation of (V) into an exact calculation of (V). Before we state this result in a theorem, we use a definition to define some terms.
Definition 106: Triple Integrals, Iterated Integration (Part I)
Let (D) be a closed, bounded region in space. Let (a) and (b) be real numbers, let (g_1(x)) and (g_2(x)) be continuous functions of (x), and let (f_1(x,y)) and (f_2(x,y)) be continuous functions of (x) and (y).
*The volume (V) of (D) is denoted by a triple integral, $$V = iiint_D dV.$$
*The iterated integral ( int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx) is evaluated as
$$int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx=int_a^bint_{g_1(x)}^{g_2(x)}left(int_{f_1(x,y)}^{f_2(x,y)} dzright) dy , dx.$$
Evaluating the above iterated integral is triple integration.
Our informal understanding of the notation (iiint_D dV) is ’sum up lots of little volumes over (D),’ analogous to our understanding of (iint_R dA) and (iint_R dm).
We now state the major theorem of this section.
theorem 125 Triple Integration (Part I)
Let (D) be a closed, bounded region in space and let (Delta D) be any subdivision of (D) into (n) rectangular solids, where the (i^text{th}) subregion (D_i) has dimensions (Delta x_itimesDelta y_itimesDelta z_i) and volume (Delta V_i).
*The volume (V) of (D) is
$$V = iiint_D dV = lim_{||Delta D||to0} sum_{i=1}^n Delta V_i = lim_{||Delta D||to0} sum_{i=1}^n Delta x_iDelta y_iDelta z_i.$$
*If (D) is defined as the region bounded by the planes (x=a) and (x=b), the cylinders (y=g_(x)) and (y=g_2(x)), and the surfaces (z=f_1(x,y)) and (z=f_2(x,y)), where (a<b), (g_1(x)leq g_2(x)) and (f_1(x,y)leq f_2(x,y)) on (D), then
$$iiint_D dV = int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx.$$
*(V) can be determined using iterated integration with other orders of integration (there are 6 total), as long as (D) is defined by the region enclosed by a pair of planes, a pair of cylinders, and a pair of surfaces.
We evaluated the area of a plane region (R) by iterated integration, where the bounds were ’from curve to curve, then from point to point.’ Theorem 125 allows us to find the volume of a space region with an iterated integral with bounds ’from surface to surface, then from curve to curve, then from point to point.’ In the iterated integral
[int_a^bint_{g_1(x)}^{g_2(x)}int_{f_1(x,y)}^{f_2(x,y)} dz , dy , dx,]
the bounds (aleq xleq b) and (g_1(x)leq yleq g_2(x)) define a region (R) in the (x)-(y) plane over which the region (D) exists in space. However, these bounds are also defining surfaces in space; (x=a) is a plane and (y=g_1(x)) is a cylinder. The combination of these 6 surfaces enclose, and define, (D).
Examples will help us understand triple integration, including integrating with various orders of integration.
Example (PageIndex{2}): Finding the volume of a space region with triple integration
Find the volume of the space region in the (1^{,st}) octant bounded by the plane (z=2-y/3-2x/3), shown in Figure 13.38(a), using the order of integration (dz , dy , dx). Set up the triple integrals that give the volume in the other 5 orders of integration.
Solution
Starting with the order of integration (dz , dy , dx), we need to first find bounds on (z). The region (D) is bounded below by the plane (z=0) (because we are restricted to the first octant) and above by (z=2-y/3-2x/3); (0leq zleq 2-y/3-2x/3).
To find the bounds on (y) and (x), we ’collapse’ the region onto the (x)-(y) plane, giving the triangle shown in Figure 13.38(b). (We know the equation of the line (y=6-2x) in two ways. First, by setting (z=0), we have (0 = 2-y/3-2x/3 Rightarrow y=6-2x). Secondly, we know this is going to be a straight line between the points ((3,0)) and ((0,6)) in the (x)-(y) plane.)
We define that region (R), in the integration order of (dy , dx), with bounds (0leq yleq 6-2x) and (0leq xleq 3). Thus the volume (V) of the region (D) is:
[begin{align*}
V &= iiint_D dV
&= int_0^3int_0^{6-2x}int_0^{2-frac 13y-frac 23x} dz dy dz
&= int_0^3int_0^{6-2x}left(int_0^{2-frac 13y-frac 23x} dzright) dy dz
&=int_0^3int_0^{6-2x}zBig|_0^{2-frac 13y-frac 23x} dy dz
&= int_0^3int_0^{6-2x}left(2-frac 13y-frac 23xright) dy dz. end{align*}]
From this step on, we are evaluating a double integral as done many times before. We skip these steps and give the final volume,
[= 6text{u}^3. ]
The order (dz , dx , dy):
Now consider the volume using the order of integration (dz , dx , dy). The bounds on (z) are the same as before, (0leq zleq 2-y/3-2x/3). Collapsing the space region on the (x)-(y) plane as shown in Figure 13.38(b), we now describe this triangle with the order of integration (dx dy). This gives bounds (0leq xleq 3-y/2) and (0leq yleq 6). Thus the volume is given by the triple integral
[V = int_0^6int_0^{3-frac12y}int_0^{2-frac13y-frac23x} dz , dx , dy.]
The order (dx , dy , dz):
Following our ’surface to surface(ldots)’ strategy, we need to determine the (x)-surfaces that bound our space region. To do so, approach the region ’from behind,’ in the direction of increasing (x). The first surface we hit as we enter the region is the (y)-(z) plane, defined by (x=0). We come out of the region at the plane (z=2-y/3-2x/3); solving for (x), we have (x= 3-y/2-3z/2). Thus the bounds on (x) are: (0leq xleq 3-y/2-3z/2).
Now collapse the space region onto the (y)-(z) plane, as shown in Figure 13.39(a). (Again, we find the equation of the line (z=2-y/3) by setting (x=0) in the equation (x=3-y/2-3z/2).) We need to find bounds on this region with the order (dy dz). The curves that bound (y) are (y=0) and (y=6-3z); the points that bound (z) are 0 and 2. Thus the triple integral giving volume is:
[begin{array}{cc}
begin{array}{c}
0leq xleq 3-y/2-3z/2
0leq yleq 6-3z
0leq zleq 2
end{array}
&
Rightarrow quad int_0^2int_0^{6-3z}int_0^{3-y/2-3z/2} dx , dy , dz.
end{array}
]
The order (dx , dz , dy):
The (x)-bounds are the same as the order above. We now consider the triangle in Figure 13.39(a) and describe it with the order (dz dy): (0leq zleq 2-y/3) and (0leq yleq 6). Thus the volume is given by:
$$begin{array}{cc}
begin{array}{c}
0leq xleq 3-y/2-3z/2
0leq zleq 2-y/3
0leq yleq 6
end{array}
&
Rightarrow quad int_0^6int_0^{2-y/3}int_0^{3-y/2-3z/2} dx , dz , dy.
end{array}
]
The order (dy dz dx):
We now need to determine the (y)-surfaces that determine our region. Approaching the space region from ’behind’ and moving in the direction of increasing (y), we first enter the region at (y=0), and exit along the plane (z= 2-y/3-2x/3). Solving for (y), this plane has equation (y = 6-2x-3z). Thus (y) has bounds (0leq yleq 6-2x-3z).
Now collapse the region onto the (x)-(z) plane, as shown in Figure 13.39(b). The curves bounding this triangle are (z=0) and (z=2-2x/3); (x) is bounded by the points (x=0) to (x=3). Thus the triple integral giving volume is:
$$begin{array}{cc}
begin{array}{c}
0leq yleq 6-2x-3z
0leq zleq 2-2x/3
0leq xleq 3
end{array}
&
Rightarrow quad int_0^3int_0^{2-2x/3}int_0^{6-2x-3z} dy dz dx.
end{array}
]
The order (dy , dx , dz):
The (y)-bounds are the same as in the order above. We now determine the bounds of the triangle in Figure 13.39(b) using the order (dy , dx , dz). (x) is bounded by (x=0) and (x=3-3z/2); (z) is bounded between (z=0) and (z=2). This leads to the triple integral:
[begin{array}{cc}
begin{array}{c}
0leq yleq 6-2x-3z
0leq xleq 3-3z/2
0leq zleq 2
end{array}
&
Rightarrow quad int_0^2int_0^{3-3z/2}int_0^{6-2x-3z} dy , dx , dz.
end{array}
]
This problem was long, but hopefully useful, demonstrating how to determine bounds with every order of integration to describe the region (D). In practice, we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best.
In the previous example, we collapsed the surface into the (x)-(y), (x)-(z), and (y)-(z) planes as we determined the ’curve to curve, point to point’ bounds of integration. Since the surface was a triangular portion of a plane, this collapsing, or projecting, was simple: the projection of a straight line in space onto a coordinate plane is a line.
The following example shows us how to do this when dealing with more complicated surfaces and curves.
Example (PageIndex{3}): Finding the projection of a curve in space onto the coordinate planes
Consider the surfaces (z=3-x^2-y^2) and (z=2y), as shown in Figure 13.40(a). The curve of their intersection is shown, along with the projection of this curve into the coordinate planes, shown dashed. Find the equations of the projections into the coordinate planes.
Solution
The two surfaces are (z=3-x^2-y^2) and (z=2y). To find where they intersect, it is natural to set them equal to each other: (3-x^2-y^2=2y). This is an implicit function of (x) and (y) that gives all points ((x,y)) in the (x)-(y) plane where the (z) values of the two surfaces are equal.
We can rewrite this implicit function by completing the square:
$$3-x^2-y^2=2y quad Rightarrow quad y^2+2y+x^2=3quad Rightarrow quad (y+1)^2+x^2=4.$$
Thus in the (x)-(y) plane the projection of the intersection is a circle with radius 2, centered at ((0,-1)).
To project onto the (x)-(z) plane, we do a similar procedure: find the (x) and (z) values where the (y) values on the surface are the same. We start by solving the equation of each surface for (y). In this particular case, it works well to actually solve for (y^2):
(z=3-x^2-y^2 quad Rightarrow quad y^2=3-x^2-z)
(z=2y quad Rightarrow quad y^2=z^2/4).
Thus we have (after again completing the square):
$$3-x^2-z = z^2/4 quad Rightarrowquad frac{(z+2)^2}{16}+frac{x^2}4=1,$$
and ellipse centered at ((0,-2)) in the (x)-(z) plane with a major axis of length 8 and a minor axis of length 4.
Finally, to project the curve of intersection into the (y)-(z) plane, we solve equation for (x). Since (z=2y) is a cylinder that lacks the variable (x), it becomes our equation of the projection in the (y)-(z) plane.
All three projections are shown in Figure 13.40(b).
Example (PageIndex{4}): Finding the volume of a space region with triple integration
Set up the triple integrals that find the volume of the space region (D) bounded by the surfaces (x^2+y^2=1), (z=0) and (z=-y), as shown in Figure 13.41(a), with the orders of integration (dz , dy , dx), (dy , dx , dz) and (dx , dz , dy).
Solution
The order (dz , dy , dx):
The region (D) is bounded below by the plane (z=0) and above by the plane (z=-y). The cylinder (x^2+y^2=1) does not offer any bounds in the (z)-direction, as that surface is parallel to the (z)-axis. Thus (0leq zleq -y).
Collapsing the region into the (x)-(y) plane, we get part of the circle with equation (x^2+y^2=1) as shown in Figure 13.41(b). As a function of (x), this half circle has equation (y=-sqrt{1-x^2}). Thus (y) is bounded below by (-sqrt{1-x^2}) and above by (y=0): (-sqrt{1-x^2}leq yleq 0). The (x) bounds of the half circle are (-1leq xleq 1). All together, the bounds of integration and triple integral are as follows:
[begin{array}{cc}
begin{array}{c}
0leq zleq -y
-sqrt{1-x^2}leq yleq 0
-1leq xleq 1
end{array}
&
Rightarrow quad int_{-1}^1int_{-sqrt{1-x^2}}^{0}int_0^{-y} dz , dy , dx.
end{array}
]
We evaluate this triple integral:
[begin{align*}
int_{-1}^1int_{-sqrt{1-x^2}}^{0}int_0^{-y} dz , dy , dx &= int_{-1}^1int_{-sqrt{1-x^2}}^{0}big(-ybig) dy , dx
&=int_{-1}^1big(-frac12y^2big)Big|_{-sqrt{1-x^2}}^{0} dx
&= int_{-1}^1 frac12big(1-x^2big) dx
&= left.left(frac12left(x-frac13x^3right)right)right|_{-1}^1
&= frac23text{units}^3.
end{align*}]
With the order (dy , dx , dz):
The region is bounded ’below’ in the (y)-direction by the surface (x^2+y^2=1 Rightarrow y=-sqrt{1-x^2}) and ’above’ by the surface (y=-z). Thus the (y) bounds are (-sqrt{1-x^2}leq yleq -z).
Collapsing the region onto the (x)-(z) plane gives the region shown in Figure 13.42(a); this half circle has equation (x^2+z^2=1). (We find this curve by solving each surface for (y^2), then setting them equal to each other. We have (y^2=1-x^2) and (y=-zRightarrow y^2=z^2). Font free download. Thus (x^2+z^2=1).) It is bounded below by (x=-sqrt{1-z^2}) and above by (x=sqrt{1-z^2}), where (z) is bounded by (0leq zleq 1). All together, we have:
[begin{array}{cc}
begin{array}{c}
-sqrt{1-x^2}leq yleq -z
-sqrt{1-z^2}leq xleq sqrt{1-z^2}
0leq zleq 1
end{array}
&
Rightarrow quad int_{0}^1int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{-sqrt{1-x^2}}^{-z} dy , dx , dz.
end{array}
]
With the order (dx , dz , dy):
(D) is bounded below by the surface (x=-sqrt{1-y^2}) and above by (sqrt{1-y^2}). We then collapse the region onto the (y)-(z) plane and get the triangle shown in Figure 13.42}(b). (The hypotenuse is the line (z=-y), just as the plane.) Thus (z) is bounded by (0leq zleq -y) and (y) is bounded by (-1leq yleq 0). This gives:
[begin{array}{cc}
begin{array}{c}
-sqrt{1-y^2}leq xleq sqrt{1-y^2}
0leq zleq -y
-1leq yleq 0
end{array}
&
Rightarrow quad int_{-1}^0int_{0}^{-y}int_{-sqrt{1-y^2}}^{sqrt{1-y^2}} dx , dz , dy.
end{array}
]
The following theorem states two things that should make ’common sense’ to us. First, using the triple integral to find volume of a region (D) should always return a positive number; we are computing volume here, not signed volume. Secondly, to compute the volume of a ’complicated’ region, we could break it up into subregions and compute the volumes of each subregion separately, summing them later to find the total volume.
THEOREM 126: Properties of Triple Integrals
Let (D) be a closed, bounded region in space, and let (D_1) and (D_2) be non-overlapping regions such that (D=D_1bigcup D_2).
*( iiint_D dV geq 0)
*( iiint_D dV = iiint_{D_1} dV + iiint_{D_2} dV.)
We use this latter property in the next example.
Example (PageIndex{5}): Finding the volume of a space region with triple integration
Find the volume of the space region (D) bounded by the coordinate planes, (z=1-x/2) and (z=1-y/4), as shown in Figure 13.43(a). Set up the triple integrals that find the volume of (D) in all 6 orders of integration.
Solution
Following the bounds--determining strategy of ’surface to surface, curve to curve, and point to point,’ we can see that the most difficult orders of integration are the two in which we integrate with respect to (z) first, for there are two ’upper’ surfaces that bound (D) in the (z)-direction. So we start by noting that we have
[0leq zleq 1-frac12x quadtext{and}quad 0leq zleq 1-frac14y.]
We now collapse the region (D) onto the (x)-(y) axis, as shown in Figure 13.43(b). The boundary of (D), the line from ((0,0,1)) to ((2,4,0)), is shown in part (b) of the figure as a dashed line;
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