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Course: Multivariable calculus>Unit 4

Lesson 7: Triple integrals

Triple integrals 1

Introduction to the triple integral. Created by Sal Khan.

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• So what would the use be of a quadruple integral?
• Quadruple integrals would be used if you are dealing with 4 dimensional space, which is something that has only ever been theorized. Theoretically, there is no limit to how many integrals you can have since mathematically, there's no limit to the number of spatial dimensions you can work with.

However, in practice, I've never seen quadruple integrals. There is very little, if any, practical application for them.
• Sal do you have any videos on evaluating double or triple integrals using plane polar,cylindrical and spherical coordinates.I can't find them anywhere
• There are articles on double integrals in Polar Coordinates in the previous section articles.
• Is there any plans to remake these videos? The entire multivariable calculus course is quite outdated and its a very fundamental topic for practically all engineering paths.
• what is a triple integral good for other then solving a basic quadratic prism?
(1 vote)
• You can calculate the volume of any 3d object.
• Is there a difference between this and volume integrals?
• Double and triple integrals are volume integrals--they are measuring the total volume of a 3-D object in the xyz-coordinate space. Analogously, "single" integrals measure the total area of a 2-D figure in the xy-coordinate plane.
Use: Function:
int -- y = f(x)
iint -- z = f(x, y)
iiint -- w = f(x, y, z)
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Note that "int" is the regular single integral, "iint" is a double integral, and "iiint" is a triple integral. While both double and triple deal with three dimensional space, the integrals are different. The triple integral measures 3-D objects while they are changing position, which brings it into the fourth dimension. Depending on the scenario, a triple integral may be a volume integral, but a double integral is, by default, a volume integral.
(1 vote)
• Is it possible to derive the formula for the volume of a sphere using triple integrals?
• I suppose you could, though it is really just make a simple problem more complicated.
http://www.wolframalpha.com/input/?i=triple+integral+calculator&f1=1&f=TripleIntegral.integrand_1&f2=z&f=TripleIntegral.intvariable1i_z&f3=-sqrt(r%5E2-x%5E2-y%5E2)&f=TripleIntegral.rangestart1i_-sqrt(r%5E2-x%5E2-y%5E2)&f4=sqrt(r%5E2-x%5E2-y%5E2)&f=TripleIntegral.rangeend1i_sqrt(r%5E2-x%5E2-y%5E2)&f5=y&f=TripleIntegral.intvariable2m_y&f6=-sqrt(r%5E2-x%5E2)&f=TripleIntegral.rangestart2m_-sqrt(r%5E2-x%5E2)&f7=sqrt(r%5E2-x%5E2)&f=TripleIntegral.rangeend2m_sqrt(r%5E2-x%5E2)&f8=x&f=TripleIntegral.intvariable3_x&f9=-r&f=TripleIntegral.rangestart3_-r&f10=r&f=TripleIntegral.rangeend3_r
• I understand the entire concept as presented in the video. However I have a lingering doubt. The triple integral was solved by imagining the 3 dimensional space, and adding up tiny cubes all over the space. In the previous section of double integration, we solved again by imaging 3 D space and adding columns of area (dxdy), height f(x,y) over the 3 D space. Is it not possible to explain double integrals with only 2 D space?
• I don't quite understand the intuition behind taking the integral in the z axis. Sal describes it in the context of a volume, but it's a one dimensional quantity. Is it something akin to the Dirac Delta function? I understand in the context of the Dirac Delta how there could be some area under the curve equal to some quantity when one axis goes to infinity, but I'm not understanding the intuition behind how there could be AUC for a one dimensional quantity that doesn't go to infinity. Can anyone help me to better understand this?
(1 vote)
• The element dxdydz in not one-dimensional. It's an infinitesimal volume element (dV, just like we could use dA in double integrals) so, in order to get the total volume, we need to add up, by integrating, the volume elements dV in each direction in turn.

The reason you can't see an area under the curve for the first integral is that, in a certain sense, there isn't one.

An integral (of ydx for the sake of argument) is not so much about finding the area under y, but rather is about adding up all of the values of y along the line x with some form of weighting to make sure it converges.

In 2D space, this comes out nicely as the area under the curve and, if we move to three dimensions, by integrating zdA we can find the volume under a curve.

This intuition doesn't help us much when we want to do triple integrals because in this case we're actually finding the 4D-hypervolume under the curve.

A more useful intuition in this case is the one of adding up all the values.

If we go back to a double integral, you remember how we integrated zdA (or zdxdy) to get the volume? Instead of thinking about z as a height above the xy plane, let's instead work in the 2D xy world where we've labelled each point in our world with a value, z.

Integrating these across some area (integrating zdA) adds up all the z's and gives us the volume we wanted.

With a triple integral, the answer is possibly a little harder to interpret so I'll work in terms of density because it's a good example.

If we have some arbitrary shape R (for a Region of space) with some arbitrary density function p and we want to find the mass M of the entire thing.

Now, the integral of dM over the entire shape should be M fairly intuitively so we need to find a way to express dM in terms of other things.

The density p is a good start because it's the only thing which has a unit containing a mass term. Density is measured in kgm^-3 so, in order to get it into a mass (kg) we're going to need to multiply it by something with units of m^3.

The obvious choice is dV=dxdydz.

This gives us that dM=pdV=pdxdydz

If we then integrate this over the entire shape R, we obtain the mass M.

Now, you'll notice, in this working, we weren't thinking about any areas under curves because that would mean we had to work in 4D in our heads and that's a nuisance so, instead we just thought of integrating as adding up a load of infinitesimals.
• Does the order of the dx,dy, and dz matter in a general scope of taking the integral. I understand that different combinations can prove to be easier but must they be in a specific/alphabetical order? Ex. If dx is first does it need to be followed by dy, then by dz or same case but we start with dy then dz then dx.
• All orders of integration will produce the same result - though as you said, some orders can be much more difficult.

The theorem: http://mathworld.wolfram.com/FubiniTheorem.html

It is a great way to check your work, eg if you did dxdydz try dxdzdy.