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Evaluating triple integrals (iterated integrals) is not that much different than evaluating double integrals. You start on the inside and work your way out, one integral at a time. The notation is a little different, of course, but it parallels double integrals.
On this page we cover triple integrals in rectangular (Cartesian) coordinates and some applications. Triple integrals in cylindrical coordinates are covered on a separate page, as well as triple integrals in spherical coordinates.
Before we get started on the details of triple integrals, let's watch a video explaining what a triple integral represents.
video by Krista King Math
The notation for triple integrals is a natural extension of double integrals. A triple integral can be expressed as
\(\displaystyle{ \iiint\limits_V {f(x,y,z) ~ dV} }\)
where V refers to a volume and dV is a differential volume. This differential volume can be expressed in six possible ways.
\(dx~dy~dz\) | \(dy~dx~dz\) | \(dz~dx~dy\) | ||
\(dx~dz~dy\) | \(dy~dz~dx\) | \(dz~dy~dx\) |
The idea with triple integrals is that we describe a volume using the three integrals and then integrate some function, \(f(x,y,z)\) over that volume. Some possible interpretations are, for \(f(x,y,z) = 1\), we are just calculating the volume itself. If \(f(x,y,z)\) is some kind of density function, we may be calculating a mass. There are many applications of triple integrals that may run across in your discipline or career. We will start with just calculating volume of a region.
Calculating volume with triple integrals extends your ability to calculate volumes to cover a larger range of volumes. With single integrals, you were mostly limited to volumes of rotation and volumes with a 'regular' or basic footprint. With double integrals, the footprint of a volume could be more complicated and you were able to calculate volumes above that footprint. With triple integrals, the volumes you will be able to calculate are more irregular and can be described with more complicated equations.
However, with more knowledge comes more difficulty. In this case, knowing what the volume looks like in three dimensions not only helps you a lot, sometimes it is necessary. You can use the free graphing utility winplot to plot three dimensional figures. It is not the best but it is free and can get you started. You can find a download link on the tools page. If you can't plot a three dimensional figure, it helps to plot the 'shadow' in each of the coordinate planes to try to get a concept of what the figure looks like.
The first key to setting up triple integrals, is to describe the region correctly. Let's start by looking at a region in the xy-plane, explained in this panel.
To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot ( used to plot these graphs; we used gimp to add labels and other graphics ). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically |
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Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow | |
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\( g(x) \leq y \leq f(x) \) |
arrow leaves through \(f(x)\) and enters through \(g(x)\) |
\( a \leq x \leq b \) |
arrow sweeps from left (\(x=a\)) to right (\(x=b\)) |
Horizontally |
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We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).
Lower Section - - This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section - - This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow | |
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lower section | |
\( a \leq x \leq G(y) \) |
arrow leaves through \(G(y)\) and enters through \(x=a\) |
\( g(a) \leq y \leq g(b) \) |
arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\)) |
upper section | |
\( a \leq x \leq F(y) \) |
arrow leaves through \(F(y)\) and enters through \(x=a\) |
\( f(b) \leq y \leq f(a) \) |
arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\)) |
Type 1 and Type 2 Regions |
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Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
video by Krista King Math
[ For some videos and practice problems dedicated to this topic, check out this page. ]
Okay, so now that you know how to describe regions in the xy-plane, you should be able to easily extend that concept to other planes.
Changing Order of Integration |
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Similar to double integrals, we can change the order of integration of triple integrals. When this is done, the limits will most likely change too. However, with triple integrals there are 6 possible ways to set up a volume integral. Compare that to only two possibilities with double integrals.
Here are three sequential videos where the presenter shows the technique of how to change the order of integration. He works a specific example (which is about the only way to show you how it works since there is no general way to do this for all cases). He does a good job of explaining the technique.
video by PatrickJMT
video by PatrickJMT
video by PatrickJMT
Application: Volume |
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In order to find the volume of the solid described as a triple integral, we just let \(f(x,y,z)=1\). This essentially means that every position \((x,y,z)\) in the volume has a weight of 1. We do not assign any units to this value. Therefore the result after integration is cubic units (whatever units x, y and z are expressed in). So, for example, if x, y and z are in meters, then the result of the integration will be the volume of the solid in cubic meters.
Okay, now for some practice problems in rectangular coordinates. There are really no new concepts to learn here. It is just a matter of putting together what you already know. The best way to do that is by working as many practice problems as you have time for.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
Instructions - Unless otherwise instructed, evaluate these integrals in the rectangular coordinate system. (Some of these problems can be successfully worked in more than one coordinate system.)
Basic Problems |
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Evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{1-x}{ \int_{0}^{1-x-y}{ ~dz} ~dy } ~dx } }\) and interpret it as a volume.
Problem Statement |
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Evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{1-x}{ \int_{0}^{1-x-y}{ ~dz} ~dy } ~dx } }\) and interpret it as a volume.
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{2}{ \int_{0}^{\sqrt{4-z^2}}{ z\sin(y)~dx } ~dz} ~dy} }\) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{2}{ \int_{0}^{\sqrt{4-z^2}}{ z\sin(y)~dx } ~dz} ~dy} }\) |
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ \int_{0}^{3}{ xyz ~dz} ~dy} ~dx} }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ \int_{0}^{3}{ xyz ~dz} ~dy} ~dx} }\)
Solution |
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video by MIP4U
close solution |
\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{y/2}{ \int_{0}^{\sqrt{y^2-x^2}}{z ~dz} ~dx} ~dy} }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{y/2}{ \int_{0}^{\sqrt{y^2-x^2}}{z ~dz} ~dx} ~dy} }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{1}^{3}{ \int_{1}^{e^2}{ \int_{1}^{1/(yz)}{ \ln(z) ~dx} ~dz} ~dy} }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{3}{ \int_{1}^{e^2}{ \int_{1}^{1/(yz)}{ \ln(z) ~dx} ~dz} ~dy} }\)
Solution |
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Note in the video that the upper limit for x is written as \(1/yz\). According to the order of operations, this means \(1/yz=(1/y)z=z/y\). However, in the video he actually means \(1/(yz)\).
video by MIP4U
close solution |
Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dz~dy~dx \).
Problem Statement |
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Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dz~dy~dx \).
Solution |
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video by MIP4U
close solution |
Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dy~dx~dz \).
Problem Statement |
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Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dy~dx~dz \).
Solution |
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video by MIP4U
close solution |
Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dx~dz~dy \).
Problem Statement |
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Express the volume bounded by \( 4z = x^2 + y^2 \) and \( z=4 \) as a triple integral with order \( dx~dz~dy \).
Solution |
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video by MIP4U
close solution |
Determine the volume of the solid bounded by \( y = -2x + 4 \) and \( z = 3 \) with \( x \geq 0, y \geq 0, z\geq 0 \).
Problem Statement |
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Determine the volume of the solid bounded by \( y = -2x + 4 \) and \( z = 3 \) with \( x \geq 0, y \geq 0, z\geq 0 \).
Solution |
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video by MIP4U
close solution |
Determine the volume of the solid bounded by \( z = y \), \( y = 1 - x^2 \) and \( z = 0 \).
Problem Statement |
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Determine the volume of the solid bounded by \( z = y \), \( y = 1 - x^2 \) and \( z = 0 \).
Solution |
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video by MIP4U
close solution |
Determine the volume of the solid in the first octant bounded by \( x + 2y + z = 6 \).
Problem Statement |
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Determine the volume of the solid in the first octant bounded by \( x + 2y + z = 6 \).
Solution |
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video by MIP4U
close solution |
Use a triple integral to find the volume of the solid bounded by \( z = 0, z = x, x = 4 - y^2 \).
Problem Statement |
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Use a triple integral to find the volume of the solid bounded by \( z = 0, z = x, x = 4 - y^2 \).
Solution |
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video by MIP4U
close solution |
Intermediate Problems |
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Sketch the solid bounded by \( x^2 + y^2 = 4 \), \( z = 1 \) and \( z = 9 - 3x - 2y \) and calculate the volume.
Problem Statement |
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Sketch the solid bounded by \( x^2 + y^2 = 4 \), \( z = 1 \) and \( z = 9 - 3x - 2y \) and calculate the volume.
Solution |
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video by Dr Chris Tisdell
close solution |
Determine the volume of the solid bounded inside \( x^2 + y^2 = 9 \) above \( z = 0 \) and below \( x + z = 4 \).
Problem Statement |
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Determine the volume of the solid bounded inside \( x^2 + y^2 = 9 \) above \( z = 0 \) and below \( x + z = 4 \).
Solution |
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video by MIP4U
close solution |
Express as a triple integral, using cartesian coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a >0 \).
Problem Statement |
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Express as a triple integral, using cartesian coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a >0 \).
Final Answer |
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\(\displaystyle{ \int_{-a}^{a}{ \int_{-\sqrt{a^2-x^2}}^{ \sqrt{a^2-x^2}}{ \int_{\sqrt{x^2+y^2}}^{a+\sqrt{a^2-(x^2+y^2)}}{ } } } }\) \(dz ~dy ~dx\) |
Problem Statement |
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Express as a triple integral, using cartesian coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a >0 \).
Solution |
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Note - This same problem is set up in cylindrical coordinates on the cylindrical coordinates page and in spherical coordinates on the spherical coordinates page. Also on the spherical coordinates page, the integral is evaluated. But you are not asked to evaluate it here, in rectangular coordinates.
video by Dr Chris Tisdell
Final Answer |
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\(\displaystyle{ \int_{-a}^{a}{ \int_{-\sqrt{a^2-x^2}}^{ \sqrt{a^2-x^2}}{ \int_{\sqrt{x^2+y^2}}^{a+\sqrt{a^2-(x^2+y^2)}}{ } } } }\) \(dz ~dy ~dx\) |
close solution |
Evaluate \( \iiint\limits_V{ z ~dV } \) where the volume V is the solid tetrahedron with vertices \((0,0,0), (5,0,0), (0,3,0), (0,0,2)\).
Problem Statement |
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Evaluate \( \iiint\limits_V{ z ~dV } \) where the volume V is the solid tetrahedron with vertices \((0,0,0), (5,0,0), (0,3,0), (0,0,2)\).
Final Answer |
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2.5 |
Problem Statement |
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Evaluate \( \iiint\limits_V{ z ~dV } \) where the volume V is the solid tetrahedron with vertices \((0,0,0), (5,0,0), (0,3,0), (0,0,2)\).
Solution |
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This problem is solved in two videos.
video by MIP4U
video by MIP4U
Final Answer |
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2.5 |
close solution |
Evaluate \( \iiint\limits_E{ 2xz ~dV } \) where \(E = \{(x,y,z) | 1 \leq x \leq 2, x \leq y \leq 2x, 0 \leq z \leq x+3y \}\).
Problem Statement |
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Evaluate \( \iiint\limits_E{ 2xz ~dV } \) where \(E = \{(x,y,z) | 1 \leq x \leq 2, x \leq y \leq 2x, 0 \leq z \leq x+3y \}\).
Solution |
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video by MIP4U
close solution |
Advanced Problems |
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Find the volume common to the two cylinders \( x^2 + y^2 \leq a^2 \) and \( x^2 + z^2 \leq a^2 \) with \( a > 0 \).
Problem Statement |
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Find the volume common to the two cylinders \( x^2 + y^2 \leq a^2 \) and \( x^2 + z^2 \leq a^2 \) with \( a > 0 \).
Solution |
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video by Dr Chris Tisdell
close solution |