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Triple Integrals
Recognize when a function of three variables is integrable over a rectangular box.
Integrable Functions of Three Variables
We can define a rectangular box \(B\) in \({ℝ}^{3}\) as \(B=\{(x,y,z)|a\le x\le b,c\le y\le d,e\le z\le f\}.\) We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval \([a,b]\) into \(l\) subintervals \([{x}_{i-1},{x}_{i}]\) of equal length \(\text{\Delta }x=\frac{b-a}{l},\) divide the interval \([c,d]\) into \(m\) subintervals \([{y}_{j-1},{y}_{j}]\) of equal length \(\text{\Delta }y=\frac{d-c}{m},\) and divide the interval \([e,f]\) into \(n\) subintervals \([{z}_{k-1},{z}_{k}]\) of equal length \(\text{\Delta }z=\frac{f-e}{n}.\) Then the rectangular box \(B\) is subdivided into \(lmn\) subboxes \({B}_{ijk}=[{x}_{i-1},{x}_{i}]\ \times \ [{y}_{j-1},{y}_{j}]\ \times \ [{z}_{k-1},{z}_{k}],\) as shown in .
For each \(i,j,\ \text{and}\ k,\) consider a sample point \(({x}_{ijk}^{*},{y}_{ijk}^{*},{z}_{ijk}^{*})\) in each sub-box \({B}_{ijk}.\) We see that its volume is \(\text{\Delta }V=\text{\Delta }x\text{\Delta }y\text{\Delta }z.\) Form the triple Riemann sum
\[\sum _{i=1}^{l}\sum _{j=1}^{m}\sum _{k=1}^{n}f({x}_{ijk}^{*},{y}_{ijk}^{*},{z}_{ijk}^{*})\text{\Delta }x\text{\Delta }y\text{\Delta }z.\]We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.
When the triple integral exists on \(B,\) the function \(f(x,y,z)\) is said to be integrable on \(B.\) Also, the triple integral exists if \(f(x,y,z)\) is continuous on \(B.\) Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, \(f\) is bounded on \(B\) and continuous except possibly on the boundary of \(B.\) The sample point \(({x}_{ijk}^{*},{y}_{ijk}^{*},{z}_{ijk}^{*})\) can be any point in the rectangular sub-box \({B}_{ijk}\) and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.
Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s thereom for triple integrals exists.
For \(a,b,c,d,e,\) and \(f\) real numbers, the iterated triple integral can be expressed in six different orderings:
Condensed — the full section is in OpenStax Calculus Volume 3.
Triple Integrals over a General Bounded Region
We now expand the definition of the triple integral to compute a triple integral over a more general bounded region \(E\) in \({ℝ}^{3}.\) The general bounded regions we will consider are of three types. First, let \(D\) be the bounded region that is a projection of \(E\) onto the \(xy\)-plane. Suppose the region \(E\) in \({ℝ}^{3}\) has the form
\[E=\{(x,y,z)|(x,y)\in D,{u}_{1}(x,y)\le z\le {u}_{2}(x,y)\}\]for two functions \(z={u}_{1}(x,y)\) and \(z={u}_{2}(x,y),\) such that \({u}_{1}(x,y)\le {u}_{2}(x,y)\) for all \((x,y)\) in \(D\) as shown in the following figure.
Similarly, we can consider a general bounded region \(D\) in the \(xy\)-plane and two functions \(y={u}_{1}(x,z)\) and \(y={u}_{2}(x,z)\) such that \({u}_{1}(x,z)\le {u}_{2}(x,z)\) for all \((x,z)\) in \(D.\) Then we can describe the solid region \(E\) in \({ℝ}^{3}\) as
\[E=\{(x,y,z)|(x,z)\in D,{u}_{1}(x,z)\le y\le {u}_{2}(x,z)\}\]where \(D\) is the projection of \(E\) onto the \(xz\)-plane and the triple integral is
\[\underset{E}{∭}f(x,y,z)dV=\underset{D}{∬}[\int _{{u}_{1}(x,z)}^{{u}_{2}(x,z)}f(x,y,z)dy]dA.\]Finally, if \(D\) is a general bounded region in the \(yz\)-plane and we have two functions \(x={u}_{1}(y,z)\) and \(x={u}_{2}(y,z)\) such that \({u}_{1}(y,z)\le {u}_{2}(y,z)\) for all \((y,z)\) in \(D,\) then the solid region \(E\) in \({ℝ}^{3}\) can be described as
\[E=\{(x,y,z)|(y,z)\in D,{u}_{1}(y,z)\le x\le {u}_{2}(y,z)\}\]where \(D\) is the projection of \(E\) onto the \(yz\)-plane and the triple integral is
\[\underset{E}{∭}f(x,y,z)dV=\underset{D}{∬}[\int _{{u}_{1}(y,z)}^{{u}_{2}(y,z)}f(x,y,z)dx]dA.\]Note that the region \(D\) in any of the planes may be of Type I or Type II as described in Double Integrals over General Regions. If \(D\) in the \(xy\)-plane is of Type I (), then
\[E=\{(x,y,z)|a\le x\le b,{g}_{1}(x)\le y\le {g}_{2}(x),{u}_{1}(x,y)\le z\le {u}_{2}(x,y)\}.\]\[\underset{E}{∭}f(x,y,z)dV=\int _{a}^{b}\ \int _{{g}_{1}(x)}^{{g}_{2}(x)}\ \int _{{u}_{1}(x,y)}^{{u}_{2}(x,y)}f(x,y,z)dz\ dy\ dx.\]\[E=\{(x,y,z)|c\le y\le d,{h}_{1}(y)\le x\le {h}_{2}(y),{u}_{1}(x,y)\le z\le {u}_{2}(x,y)\}.\]\[\underset{E}{∭}f(x,y,z)dV={\int }_{y=c}^{y=d}{\int }_{x={h}_{1}(y)}^{x={h}_{2}(y)}{\int }_{z={u}_{1}(x,y)}^{z={u}_{2}(x,y)}f(x,y,z)dz\ dx\ dy.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Changing the Order of Integration
As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.
Condensed — the full section is in OpenStax Calculus Volume 3.
Average Value of a Function of Three Variables
Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.
Example
Try it.
The temperature at a point \((x,y,z)\) of a solid \(E\) bounded by the coordinate planes and the plane \(x+y+z=1\) is \(T(x,y,z)=(xy+8z+20)\text{^{\circ}}\text{C}\text{.}\) Find the average temperature over the solid.
Solution
Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane \(x+y+z=1\) has intercepts \((1,0,0),(0,1,0),\) and \((0,0,1).\) The region \(E\) looks like
\[E=\{(x,y,z)|0\le x\le 1,0\le y\le 1-x,0\le z\le 1-x-y\}.\]Hence the triple integral of the temperature is
\[\underset{E}{∭}f(x,y,z)dV=\int _{x=0}^{x=1}\ \int _{y=0}^{y=1-x}\ \int _{z=0}^{z=1-x-y}(xy+8z+20)dz\ dy\ dx=\frac{147}{40}.\]The volume evaluation is \(V(E)=\underset{E}{∭}1dV=\int _{x=0}^{x=1}\ \int _{y=0}^{y=1-x}\ \int _{z=0}^{z=1-x-y}1dz\ dy\ dx=\frac{1}{6}.\)
Hence the average value is \({T}_{\text{ave}}=\frac{147\text{/}40}{1\text{/}6}=\frac{6(147)}{40}=\frac{441}{20}\) degrees Celsius.
Key Concepts
- To compute a triple integral we use Fubini’s theorem, which states that if \(f(x,y,z)\) is continuous on a rectangular box \(B=[a,b]\ \times \ [c,d]\ \times \ [e,f],\) then
\[\underset{B}{∭}f(x,y,z)dV=\int _{e}^{f}\ \int _{c}^{d}\ \int _{a}^{b}f(x,y,z)dx\ dy\ dz\]
and is also equal to any of the other five possible orderings for the iterated triple integral. - To compute the volume of a general solid bounded region \(E\) we use the triple integral
\[V(E)=\underset{E}{∭}1dV.\] - Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
- To compute the average value of a function over a general three-dimensional region, we use
\[{f}_{\text{ave}}=\frac{1}{V(E)}\underset{E}{∭}f(x,y,z)dV.\]
Key Equations
| Triple integral | \(\underset{l,m,n\to \infty }{\text{lim}}\sum _{i=1}^{l}\sum _{j=1}^{m}\sum _{k=1}^{n}f({x}_{ijk}^{*},{y}_{ijk}^{*},{z}_{ijk}^{*})\text{\Delta }x\text{\Delta }y\text{\Delta }z=\underset{B}{∭}f(x,y,z)dV\) |
Triple Integrals
In the following exercises, evaluate the triple integrals over the rectangular solid box \(B.\)
In the following exercises, change the order of integration by integrating first with respect to \(z,\) then \(x,\) then \(y.\)
In the following exercises, evaluate the triple integrals over the bounded region \(E=\{(x,y,z)|a\le x\le b,{h}_{1}(x)\le y\le {h}_{2}(x),e\le z\le f\}.\)
In the following exercises, evaluate the triple integrals over the indicated bounded region \(E.\)
In the following exercises, evaluate the triple integrals over the bounded region \(E\) of the form \(E=\{(x,y,z)|{g}_{1}(y)\le x\le {g}_{2}(y),c\le y\le d,e\le z\le f\}.\)
In the following exercises, evaluate the triple integrals over the bounded region
\[E=\{(x,y,z)|{g}_{1}(y)\le x\le {g}_{2}(y),c\le y\le d,{u}_{1}(x,y)\le z\le {u}_{2}(x,y)\}.\]In the following exercises, evaluate the triple integrals over the bounded region
Condensed — the full section is in OpenStax Calculus Volume 3.
Triple Integrals
This section extends the tools of to functions of three variables to define triple integrals. While these ideas are geometrically challenging to students (finding bounds of integration and visualization of volumes in three dimensions), many students understand the algebraic mechanics of evaluating iterated integrals of this section quickly and without much prompting.
Introduction
In this chapter, we defined the double integral of a continuous function \(f = f(x,y)\) over a rectangle \(R = [a,b] \times [c,d]\) as a limit of a double Riemann sum, which paralleled the definition of a single-variable integral for a function \(g = g(x)\) on an interval \([a,b]\). We have also repeatedly emphasized the interpretations and applications of the double integral as stated in . These ideas will naturally extend to functions with more than two variables, but the geometric elements of these extensions become harder to visualize. In the Preview Activity below, we will setup a Riemann sum as an approximation to integrating a function of three variables (step 1 of a classic calculus approach).
In the Preview Activity, we completed step 1 and justified step 2 of the, classic calculus approach toward integrating a density function in three dimensions. The generalization of our work in the previous part of this chapter on double integrals works much as you would expect, so we be brief in our statements of these generalizations and will spend the rest of this section getting used to the mechanics and geometric arguments that are typical in these types of problems.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Triple Riemann Sums and Triple Integrals
Through the application of a mass density distribution over a three-dimensional solid block of granite, Preview Activity suggests a natural generalization from double Riemann sums of functions of two variables to triple Riemann sums of functions of three variables. In the same way, we can generalize from double integrals to triple integrals. By simply adding a \(z\)-coordinate to our earlier work, we can define both a triple Riemann sum and the corresponding triple integral.
If \(f(x,y,z)\) represents the density of a material in the box \(B\), then as we saw in Preview Activity, the triple Riemann sum approximates the total mass of material in the box \(B\) as step 1 of our classic calculus approach. In order to find the exact mass of the box, we will need to let the number of sub-boxes increase without bound (in other words, let \(m\), \(n\), and \(\ell\) go to infinity). When we do this, the finite sum of the mass approximations becomes the exact mass of the solid \(B\). More generally, we have the following definition of the triple integral.
In the above description, we used equally spaced division of each of the coordinates to separate the region of integration into pieces. As the number of pieces increases (\(m\), \(n\), and \(\ell\) go to infinity), the volume of each piece (\(\Delta V\)) will go to zero. A more general treatment of this process will show that the pieces do not need to be the same size, but the size of all of the pieces must approach zero in the limit used for the triple Riemann sum.
The next activity asks you to find the mass of a solid region that does not have a base lying in the \(xy\)-plane.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Evaluate the triple integral \({\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}{\int }_{x=-1}^{x=5}(x+y{z}^{2})dx\ dy\ dz.\)
Bonisa impendulo
The order of integration is specified in the problem, so integrate with respect to \(x\) first, then y, and then \(z.\)
\[\begin{array}{llll} \\ \\ \\ \\ {\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}{\int }_{x=-1}^{x=5}(x+y{z}^{2})dx\ dy\ dz & & & \\ ={\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}[{\frac{{x}^{2}}{2}+xy{z}^{2}|}_{x=-1}^{x=5}]dy\ dz & & & \text{Integrate with respect to}\ x. \\ ={\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}[12+6y{z}^{2}]dy\ dz & & & \text{Evaluate.} \\ ={\int }_{z=0}^{z=1}[{12y+6\frac{{y}^{2}}{2}{z}^{2}|}_{y=2}^{y=4}]dz & & & \text{Integrate with respect to}\ y. \\ ={\int }_{z=0}^{z=1}[24+36{z}^{2}]dz & & & \text{Evaluate.} \\ ={[24z+36\frac{{z}^{3}}{3}]}_{z=0}^{z=1}=36. & & & \text{Integrate with respect to}\ z.\end{array}\] -
Evaluate the triple integral \(\underset{B}{∭}{x}^{2}yz\ dV\) where \(B=\{(x,y,z)|-2\le x\le 1,0\le y\le 3,1\le z\le 5\}\) as shown in the following figure.
Bonisa impendulo
The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate y first, then x, and then z.
\(\begin{array}{ll}\underset{B}{∭}{x}^{2}yz\ dV & =\int _{1}^{5}\ \int _{-2}^{1}\ \int _{0}^{3}[{x}^{2}yz]dy\ dx\ dz=\int _{1}^{5}\ \int _{-2}^{1}[{{x}^{2}\frac{{y}^{2}}{2}z|}_{0}^{3}]dx\ dz \\ & =\int _{1}^{5}\ \int _{-2}^{1}\frac{9}{2}{x}^{2}z\ dx\ dz=\int _{1}^{5}[{\frac{9}{2}\frac{{x}^{3}}{3}z|}_{-2}^{1}]dz=\int _{1}^{5}\frac{27}{2}z\ dz={\frac{27}{2}\frac{{z}^{2}}{2}|}_{1}^{5}=162.\end{array}\)
Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to \(x\) first, then \(z,\) and then \(y.\)
\[\begin{array}{ll}\underset{B}{∭}{x}^{2}yz\ dV & =\int _{0}^{3}\ \int _{1}^{5}\ \int _{-2}^{1}[{x}^{2}yz]dx\ dz\ dy=\int _{0}^{3}\ \int _{1}^{5}[{\frac{{x}^{3}}{3}yz|}_{-2}^{1}]dz\ dy \\ & =\int _{0}^{3}\ \int _{1}^{5}3yz\ dz\ dy=\int _{0}^{3}[{3y\frac{{z}^{2}}{2}|}_{1}^{5}]dy=\int _{0}^{3}36y\ dy={36\frac{{y}^{2}}{2}|}_{0}^{3}=18(9-0)=162.\end{array}\] -
Evaluate the triple integral \(\underset{B}{∭}z\ \text{sin}\ x\ \text{cos}\ y\ dV\) where \(B=\{(x,y,z)|0\le x\le \pi ,\frac{3\pi }{2}\le y\le 2\pi ,1\le z\le 3\}.\)
Bonisa impendulo
\(\underset{B}{∭}z\ \text{sin}\ x\ \text{cos}\ y\ dV=8\)
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Evaluate the triple integral of the function \(f(x,y,z)=5x-3y\) over the solid tetrahedron bounded by the planes \(x=0,y=0,z=0,\) and \(x+y+z=1.\)
Bonisa impendulo
shows the solid tetrahedron \(E\) and its projection \(D\) on the \(xy\)-plane.
We can describe the solid region tetrahedron as
\[E=\{(x,y,z)|0\le x\le 1,0\le y\le 1-x,0\le z\le 1-x-y\}.\]Hence, the triple integral is
\[\underset{E}{∭}f(x,y,z)dV={\int }_{x=0}^{x=1}{\int }_{y=0}^{y=1-x}{\int }_{z=0}^{z=1-x-y}(5x-3y)dz\ dy\ dx.\]To simplify the calculation, first evaluate the integral \({\int }_{z=0}^{z=1-x-y}(5x-3y)dz.\) We have
\[{\int }_{z=0}^{z=1-x-y}(5x-3y)dz=(5x-3y)(1-x-y).\]Now evaluate the integral \({\int }_{y=0}^{y=1-x}(5x-3y)(1-x-y)dy,\) obtaining
\[{\int }_{y=0}^{y=1-x}(5x-3y)(1-x-y)dy=\frac{1}{2}{(x-1)}^{2}(6x-1).\]Finally, evaluate
\[{\int }_{x=0}^{x=1}\frac{1}{2}{(x-1)}^{2}(6x-1)dx=\frac{1}{12}.\]Putting it all together, we have
\[\underset{E}{∭}f(x,y,z)dV={\int }_{x=0}^{x=1}{\int }_{y=0}^{y=1-x}{\int }_{z=0}^{z=1-x-y}(5x-3y)dz\ dy\ dx=\frac{1}{12}.\] -
Find the volume of a right pyramid that has the square base in the \(xy\)-plane \([-1,1]\ \times \ [-1,1]\) and vertex at the point \((0,0,1)\) as shown in the following figure.
Bonisa impendulo
In this pyramid the value of \(z\) changes from \(0\ \text{to}\ 1,\) and at each height \(z,\) the cross section of the pyramid for any value of \(z\) is the square \([-1+z,1-z]\ \times \ [-1+z,1-z].\) Hence, the volume of the pyramid is \(\underset{E}{∭}1dV\) where
\[E=\{(x,y,z)|0\le z\le 1,-1+z\le y\le 1-z,-1+z\le x\le 1-z\}.\]Thus, we have
\[\underset{E}{∭}1dV={\int }_{z=0}^{z=1}{\int }_{y=-1+z}^{y=1-z}{\int }_{x=-1+z}^{x=1-z}1dx\ dy\ dz={\int }_{z=0}^{z=1}{\int }_{y=-1+z}^{y=1-z}(2-2z)dy\ dz={\int }_{z=0}^{z=1}{(2-2z)}^{2}dz=\frac{4}{3}.\]Hence, the volume of the pyramid is \(\frac{4}{3}\) cubic units.
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Consider the solid sphere \(E=\{(x,y,z)|{x}^{2}+{y}^{2}+{z}^{2}\le 9\}.\) Write the triple integral \(\underset{E}{∭}f(x,y,z)dV\) for an arbitrary function \(f\) as an iterated integral. Then evaluate this triple integral with \(f(x,y,z)=1.\) Notice that this gives the volume of a sphere using a triple integral.
Bonisa impendulo
\(\underset{E}{∭}1dV={\int }_{x=-3}^{x=3}{\int }_{y=\text{-}\sqrt{9-{x}^{2}}}^{y=\sqrt{9-{x}^{2}}}{\int }_{z=\text{-}\sqrt{9-{x}^{2}-{y}^{2}}}^{z=\sqrt{9-{x}^{2}-{y}^{2}}}1dz\ dy\ dx=36\pi .\)
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Consider the iterated integral
\[\int _{x=0}^{x=1}\ \int _{y=0}^{y={x}^{2}}\ \int _{z=0}^{z={{y}^{2}}^{}}f(x,y,z)dz\ dy\ dx.\]The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to x, then z, and then \(y.\) Verify that the value of the integral is the same if we let \(f(x,y,z)=xyz.\)
Bonisa impendulo
The best way to do this is to sketch the region \(E\) and its projections onto each of the three coordinate planes. Thus, let
\[E=\{(x,y,z)|0\le x\le 1,0\le y\le {x}^{2},0\le z\le {y}^{2}\}.\]and
\[\int _{x=0}^{x=1}\ \int _{y=0}^{y={x}^{2}}\ \int _{z=0}^{z={y}^{2}}f(x,y,z)dz\ dy\ dx=\underset{E}{∭}f(x,y,z)dV.\]We need to express this triple integral as
\[\int _{y=c}^{y=d}\ \int _{z={v}_{1}(y)}^{z={v}_{2}(y)}\ \int _{x={u}_{1}(y,z)}^{x={u}_{2}(y,z)}f(x,y,z)dx\ dz\ dy.\]Knowing the region \(E\) we can draw the following projections ():
on the \(xy\)-plane is \({D}_{1}=\{(x,y)|0\le x\le 1,0\le y\le {x}^{2}\}=\{(x,y)|0\le y\le 1,\sqrt{y}\le x\le 1\},\)
on the \(yz\)-plane is \({D}_{2}=\{(y,z)|0\le y\le 1,0\le z\le {y}^{2}\},\) and
on the \(xz\)-plane is \({D}_{3}=\{(x,z)|0\le x\le 1,0\le z\le {x}^{4}\}.\)
Now we can describe the same region \(E\) as \(\{(x,y,z)|0\le y\le 1,0\le z\le {y}^{2},\sqrt{y}\le x\le 1\},\) and consequently, the triple integral becomes
\[\int _{y=c}^{y=d}\ \int _{z={v}_{1}(y)}^{z={v}_{2}(y)}\ \int _{x={u}_{1}(y,z)}^{x={u}_{2}(y,z)}f(x,y,z)dx\ dz\ dy=\int _{y=0}^{y=1}\ \int _{z=0}^{z={y}^{2}}\ \int _{x=\sqrt{y}}^{x=1}f(x,y,z)dx\ dz\ dy.\]Now assume that \(f(x,y,z)=xyz\) in each of the integrals. Then we have
\[\begin{array}{l} \\ \\ \\ \\ \\ \int _{x=0}^{x=1}\ \int _{y=0}^{y={x}^{2}}\ \int _{z=0}^{z={y}^{2}}xyz\ dz\ dy\ dx \\ =\int _{x=0}^{x=1}\ \int _{y=0}^{y={x}^{2}}[{xy\frac{{z}^{2}}{2}|}_{z=0}^{z={y}^{2}}]dy\ dx=\int _{x=0}^{x=1}\ \int _{y=0}^{y={x}^{2}}(x\frac{{y}^{5}}{2})dy\ dx=\int _{x=0}^{x=1}[{x\frac{{y}^{6}}{12}|}_{y=0}^{y={x}^{2}}]dx=\int _{x=0}^{x=1}\frac{{x}^{13}}{12}dx=\frac{1}{168}, \\ \int _{y=0}^{y=1}\ \int _{z=0}^{z={y}^{2}}\ \int _{x=\sqrt{y}}^{x=1}xyz\ dx\ dz\ dy \\ =\int _{y=0}^{y=1}\ \int _{z=0}^{z={y}^{2}}[{yz\frac{{x}^{2}}{2}|}_{\sqrt{y}}^{1}]dz\ dy \\ =\int _{y=0}^{y=1}\ \int _{z=0}^{z={y}^{2}}(\frac{yz}{2}-\frac{{y}^{2}z}{2})dz\ dy=\int _{y=0}^{y=1}[{\frac{y{z}^{2}}{4}-\frac{{y}^{2}{z}^{2}}{4}|}_{z=0}^{z={y}^{2}}]dy=\int _{y=0}^{y=1}(\frac{{y}^{5}}{4}-\frac{{y}^{6}}{4})dy=\frac{1}{168}.\end{array}\]The answers match.
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Write five different iterated integrals equal to the given integral
\[\int _{z=0}^{z=4}\ \int _{y=0}^{y=4-z}\ \int _{x=0}^{x=\sqrt{y}}f(x,y,z)dx\ dy\ dz.\]Bonisa impendulo
(i) \(\int _{z=0}^{z=4}\ \int _{x=0}^{x=\sqrt{4-z}}\ \int _{y={x}^{2}}^{y=4-z}f(x,y,z)dy\ dx\ dz,\) (ii) \(\int _{y=0}^{y=4}\ \int _{z=0}^{z=4-y}\ \int _{x=0}^{x=\sqrt{y}}f(x,y,z)dx\ dz\ dy,\) (iii) \(\int _{y=0}^{y=4}\ \int _{x=0}^{x=\sqrt{y}}\ \int _{z=0}^{z=4-y}f(x,y,z)dz\ dx\ dy,\) (iv) \(\int _{x=0}^{x=2}\ \int _{y={x}^{2}}^{y=4}\ \int _{z=0}^{z=4-y}f(x,y,z)dz\ dy\ dx,\) (v) \(\int _{x=0}^{x=2}\ \int _{z=0}^{z=4-{x}^{2}}\ \int _{y={x}^{2}}^{y=4-z}f(x,y,z)dy\ dz\ dx\)
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Evaluate the triple integral \(\underset{E}{∭}\sqrt{{x}^{2}+{z}^{2}}dV,\) where \(E\) is the region bounded by the paraboloid \(y={x}^{2}+{z}^{2}\) () and the plane \(y=4.\)
Bonisa impendulo
The projection of the solid region \(E\) onto the \(xy\)-plane is the region bounded above by \(y=4\) and below by the parabola \(y={x}^{2}\) as shown.
Thus, we have
\[E=\{(x,y,z)|-2\le x\le 2,{x}^{2}\le y\le 4,\text{-}\sqrt{y-{x}^{2}}\le z\le \sqrt{y-{x}^{2}}\}.\]The triple integral becomes
\[\underset{E}{∭}\sqrt{{x}^{2}+{z}^{2}}dV=\int _{x=-2}^{x=2}\ \int _{y={x}^{2}}^{y=4}\ \int _{z=\text{-}\sqrt{y-{x}^{2}}}^{z=\sqrt{y-{x}^{2}}}\sqrt{{x}^{2}+{z}^{2}}dz\ dy\ dx.\]This expression is difficult to compute, so consider the projection of \(E\) onto the \(xz\)-plane. This is a circular disc \({x}^{2}+{z}^{2}\le 4.\) So we obtain
\[\underset{E}{∭}\sqrt{{x}^{2}+{z}^{2}}dV=\int _{x=-2}^{x=2}\ \int _{y={x}^{2}}^{y=4}\ \int _{z=\text{-}\sqrt{y-{x}^{2}}}^{z=\sqrt{y-{x}^{2}}}\sqrt{{x}^{2}+{z}^{2}}dz\ dy\ dx=\int _{x=-2}^{x=2}\ \int _{z=\text{-}\sqrt{4-{x}^{2}}}^{z=\sqrt{4-{x}^{2}}}\ \int _{y={x}^{2}+{z}^{2}}^{y=4}\sqrt{{x}^{2}+{z}^{2}}dy\ dz\ dx.\]Here the order of integration changes from being first with respect to \(z,\) then \(y,\) and then \(x\) to being first with respect to \(y,\) then to \(z,\) and then to \(x.\) It will soon be clear how this change can be beneficial for computation. We have
\[\int _{x=-2}^{x=2}\ \int _{z=\text{-}\sqrt{4-{x}^{2}}}^{z=\sqrt{4-{x}^{2}}}\ \int _{y={x}^{2}+{z}^{2}}^{y=4}\sqrt{{x}^{2}+{z}^{2}}dy\ dz\ dx=\int _{x=-2}^{x=2}\ \int _{z=\text{-}\sqrt{4-{x}^{2}}}^{z=\sqrt{4-{x}^{2}}}(4-{x}^{2}-{z}^{2})\sqrt{{x}^{2}+{z}^{2}}dz\ dx.\]Now use the polar substitution \(x=r\ \text{cos}\ \theta ,z=r\ \text{sin}\ \theta ,\) and \(dz\ dx=r\ dr\ d\theta\) in the \(xz\)-plane. This is essentially the same thing as when we used polar coordinates in the \(xy\)-plane, except we are replacing \(y\) by \(z.\) Consequently the limits of integration change and we have, by using \({r}^{2}={x}^{2}+{z}^{2},\)
\[\begin{array}{ll}\int _{x=-2}^{x=2}\ \int _{z=\text{-}\sqrt{4-{x}^{2}}}^{z=\sqrt{4-{x}^{2}}}(4-{x}^{2}-{z}^{2})\sqrt{{x}^{2}+{z}^{2}}dz\ dx & =\int _{\theta =0}^{\theta =2\pi }\ \int _{r=0}^{r=2}(4-{r}^{2})rr\ dr\ d\theta \\ & =\int _{0}^{2\pi }[{\frac{4{r}^{3}}{3}-\frac{{r}^{5}}{5}|}_{0}^{2}]d\theta =\int _{0}^{2\pi }\frac{64}{15}d\theta =\frac{128\pi }{15}.\end{array}\] -
The temperature at a point \((x,y,z)\) of a solid \(E\) bounded by the coordinate planes and the plane \(x+y+z=1\) is \(T(x,y,z)=(xy+8z+20)\text{^{\circ}}\text{C}\text{.}\) Find the average temperature over the solid.
Bonisa impendulo
Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane \(x+y+z=1\) has intercepts \((1,0,0),(0,1,0),\) and \((0,0,1).\) The region \(E\) looks like
\[E=\{(x,y,z)|0\le x\le 1,0\le y\le 1-x,0\le z\le 1-x-y\}.\]Hence the triple integral of the temperature is
\[\underset{E}{∭}f(x,y,z)dV=\int _{x=0}^{x=1}\ \int _{y=0}^{y=1-x}\ \int _{z=0}^{z=1-x-y}(xy+8z+20)dz\ dy\ dx=\frac{147}{40}.\]The volume evaluation is \(V(E)=\underset{E}{∭}1dV=\int _{x=0}^{x=1}\ \int _{y=0}^{y=1-x}\ \int _{z=0}^{z=1-x-y}1dz\ dy\ dx=\frac{1}{6}.\)
Hence the average value is \({T}_{\text{ave}}=\frac{147\text{/}40}{1\text{/}6}=\frac{6(147)}{40}=\frac{441}{20}\) degrees Celsius.
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Find the average value of the function \(f(x,y,z)=xyz\) over the cube with sides of length \(4\) units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.
Bonisa impendulo
\({f}_{\text{ave}}=8\)
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\(\underset{B}{∭}(2x+3{y}^{2}+4{z}^{3})dV,\) where \(B=\{(x,y,z)|0\le x\le 1,0\le y\le 2,0\le z\le 3\}\)
Bonisa impendulo
\(192\)
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\(\underset{B}{∭}(xy+yz+xz)dV,\) where \(B=\{(x,y,z)|1\le x\le 2,0\le y\le 2,1\le z\le 3\}\)
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\(\underset{B}{∭}(x\ \text{cos}\ y+z)dV,\) where \(B=\{(x,y,z)|0\le x\le 1,0\le y\le \pi ,-1\le z\le 1\}\)
Bonisa impendulo
\(0\)
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\(\underset{B}{∭}(z\ \text{sin}\ x+{y}^{2})dV,\) where \(B=\{(x,y,z)|0\le x\le \pi ,0\le y\le 1,-1\le z\le 2\}\)
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\(\int _{0}^{1}\ \int _{1}^{2}\ \int _{2}^{3}({x}^{2}+\text{ln}\ y+z)dx\ dy\ dz\)
Bonisa impendulo
\(\int _{1}^{2}\ \int _{2}^{3}\ \int _{0}^{1}({x}^{2}+\text{ln}\ y+z)dz\ dx\ dy=\frac{35}{6}+2\ \text{ln}\ 2\)
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\(\int _{-1}^{1}\ \int _{0}^{3}\ \int _{0}^{1}(z{e}^{x}+2y)dx\ dy\ dz\)
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\(\int _{-1}^{2}\ \int _{1}^{3}\ \int _{0}^{4}({x}^{2}z+\frac{1}{y})dx\ dy\ dz\)
Bonisa impendulo
\(\int _{1}^{3}\ \int _{0}^{4}\ \int _{-1}^{2}({x}^{2}z+\frac{1}{y})dz\ dx\ dy=64+12\ \text{ln}\ 3\)
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\(\int _{1}^{2}\ \int _{-2}^{-1}\ \int _{0}^{1}\frac{x+y}{z}dx\ dy\ dz\)
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Let \(F,G,\ \text{and}\ H\) be continuous functions on \([a,b],[c,d],\) and \([e,f],\) respectively, where \(a,b,c,d,e,\ \text{and}\ f\) are real numbers such that \(a
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Let \(F,G,\ \text{and}\ H\) be differential functions on \([a,b],[c,d],\) and \([e,f],\) respectively, where \(a,b,c,d,e,\ \text{and}\ f\) are real numbers such that \(a
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\(\underset{E}{∭}(2x+5y+7z)dV,\) where \(E=\{(x,y,z)|0\le x\le 1,0\le y\le -x+1,1\le z\le 2\}\)
Bonisa impendulo
\(\frac{77}{12}\)
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\(\underset{E}{∭}(y\ \text{ln}\ x+z)dV,\) where \(E=\{(x,y,z)|1\le x\le e,0\le y\le \text{ln}\ x,0\le z\le 1\}\)
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\(\underset{E}{∭}(\text{sin}\ x+\text{sin}\ y)dV,\) where \(E=\{(x,y,z)|0\le x\le \frac{\pi }{2},\text{-}\text{cos}\ x\le y\le \text{cos}\ x,-1\le z\le 1\}\)
Bonisa impendulo
\(2\)
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\(\underset{E}{∭}(xy+yz+xz)dV,\) where \(E=\{(x,y,z)|0\le x\le 1,\text{-}{x}^{2}\le y\le {x}^{2},0\le z\le 1\}\)
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\(\underset{E}{∭}(x+2yz)dV,\) where \(E=\{(x,y,z)|0\le x\le 1,0\le y\le x,0\le z\le 5-x-y\}\)
Bonisa impendulo
\(\frac{439}{120}\)
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\(\underset{E}{∭}({x}^{3}+{y}^{3}+{z}^{3})dV,\) where \(E=\{(x,y,z)|0\le x\le 2,0\le y\le 2x,0\le z\le 4-x-y\}\)
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\(\underset{E}{∭}y\ dV,\) where \(E=\{(x,y,z)|-1\le x\le 1,\text{-}\sqrt{1-{x}^{2}}\le y\le \sqrt{1-{x}^{2}},0\le z\le 1-{x}^{2}-{y}^{2}\}\)
Bonisa impendulo
\(0\)
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\(\underset{E}{∭}x\ dV,\) where \(E=\{(x,y,z)|-2\le x\le 2,-\sqrt{4-{x}^{2}}\le y\le \sqrt{4-{x}^{2}},0\le z\le 4-{x}^{2}-{y}^{2}\}\)
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\(\underset{E}{∭}{x}^{2}dV,\) where \(E=\{(x,y,z)|1-{y}^{2}\le x\le {y}^{2}-1,-1\le y\le 1,1\le z\le 2\}\)
Bonisa impendulo
\(-\frac{64}{105}\)
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\(\underset{E}{∭}(\text{sin}\ x+y)dV,\) where \(E=\{(x,y,z)|-{y}^{4}\le x\le {y}^{4},0\le y\le 2,0\le z\le 4\}\)
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\(\underset{E}{∭}(x-yz)dV,\) where \(E=\{(x,y,z)|-{y}^{6}\le x\le \sqrt{y},0\le y\le 1,-1\le z\le 1\}\)
Bonisa impendulo
\(\frac{11}{26}\)
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\(\underset{E}{∭}zdV,\) where \(E=\{(x,y,z)|2-2y\le x\le 2+\sqrt{y},0\le y\le 1,2\le z\le 3\}\)
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\(\underset{E}{∭}zdV,\) where \(E=\{(x,y,z)|-y\le x\le y,0\le y\le 1,0\le z\le 1-{x}^{4}-{y}^{4}\}\)
Bonisa impendulo
\(\frac{113}{450}\)
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\(\underset{E}{∭}(xz+1)dV,\) where \(E=\{(x,y,z)|0\le x\le \sqrt{y},0\le y\le 2,0\le z\le 1-{x}^{2}-{y}^{2}\}\)
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\(\underset{E}{∭}(x-z)dV,\) where \(E=\{(x,y,z)|-\sqrt{1-{y}^{2}}\le x\le 0,0\le y\le \frac{1}{2},0\le z\le 1-{x}^{2}-{y}^{2}\}\)
Bonisa impendulo
\(\frac{-609-216\sqrt{3}-80\pi }{5760}\approx -0.21431\)
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\(\underset{E}{∭}(x+y)dV,\) where \(E=\{(x,y,z)|0\le x\le \sqrt{1-{y}^{2}},0\le y\le 1,0\le z\le 1-x\}\)
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\(\underset{D}{∬}(\int _{1}^{2}(x+z)dz)dA,\) where \(D=\{(x,y)|{x}^{2}+{y}^{2}\le 1\}\)
Bonisa impendulo
\(\frac{3\pi }{2}\)
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\(\underset{D}{∬}(\int _{1}^{3}x(z+1)dz)dA,\) where \(D=\{(x,y)|{x}^{2}-{y}^{2}\ge 1,1\le x\le \sqrt{5}\}\)
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\(\underset{D}{∬}(\int _{0}^{10-x-y}(x+2z)dz)dA,\) where \(D=\{(x,y)|y\ge 0,x\ge 0,x+y\le 10\}\)
Bonisa impendulo
\(1250\)
Symbols used here
Add a_k for k = 1 up to n.
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
x belongs to A; every element of A is in B.
Instantaneous rate of change; slope of the graph.
i² = −1.
1/360 of a full turn. 180° = π radians.
Inequalities that allow equality; < and > exclude it.
Least upper bound, greatest lower bound.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Triple Integrals
- Recognize when a function of three variables is integrable over a rectangular box.
- Evaluate a triple integral by expressing it as an iterated integral.
- Recognize when a function of three variables is integrable over a closed and bounded region.
- Simplify a calculation by changing the order of integration of a triple integral.
- Calculate the average value of a function of three variables.
- To compute a triple integral we use Fubini’s theorem, which states that if
- To compute the volume of a general solid bounded region
- Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Zama wena
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Okuningi Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems