$$$$

15  Triple Integrals

Triple integrals follow a very similar general theory to double integrals: starting with a function $f(x,y,z)$ on ${\mathbb{R}}^3$, we define the integral over a region $E$ by breaking that region into small cubical volumes of size $dV$ and building a 3-dimensional riemann sum. Taking the limit gives the triple integral, or

$${\iiint }_{E}fdV$$

To evaluate such an expression, we need to break the integral into slices, and evaluate them one at a time. Such slicing relies on understanding the volume element in three dimensions, which is the volume of an infinitesimal box. Since a box’s volume is given by length width and height, we can express $dV$ as a product of three infinitesimal lengths:

$$dV=dxdydz$$

This lets us separate the triple integral into three consecutive integrals: first dx, then dy then dz. Or, because the order of multiplication doesn’t matter, we could do the integral in any of the other six possible orders $$dxdydz=dxdzdy=dydxdz$$ $$=dydzdx=dzdxdy=dzdydx$$

15.1 Different Bounds:

Because we are going to evaluate a triple integral as three iterated integrals, we can reuse a lot of what we learned about double integrals along the process. In particular, we can think of the process of computing a triple integral as first choosing one of the directions to integrate, and then treating the two remaining directions as a double integral over a region parameterizing all the slices (if we integrate $z$, each slice has a different $x,y$ coordinate, so the region parameterizing the slices is a side of our 3D region in the $xy$ direction).

$${\iiint }_{E}fdV={\iint }_R\left(\mathrm{Slice}\right)dA={\iint }_R\left({\int }_{z_{\mathrm{start}}}^{z_{\mathrm{stop}}}fdz\right)dA$$

This allows us to think about triple integrals as not a new thing, but just adding one more direction to a process we already understand well.

15.1.1 Boxes

When the domain $E\subset {\mathbb{R}}^3$ is a coordinate box, described as $$E=\{(x,y,z)\mid a\le x\le bc\le y\le de\le z\le f\}$$

This triple integral splits into an iterated integral with constant bounds:

$${\iiint }_{E}FdV={\int }_a^b{\int }_c^d{\int }_e^{f}Fdzdydx$$

This integral could be done in any of the six possible orders, as all the bounds are constants, no order will be easier or harder than any other.

Example 15.1 $${\iiint }_{E}x{e}^{y}dVE=\{1\le x\le 2,0\le y\le 1,2\le z\le 5\}$$

First we choose an order: say we integrate $dz$ first. These bounds go from $2$ to $5$ so the integral along our slice is $$\mathrm{Slice}={\int }_2^{5}x{e}^{y}dz=x{e}^{y}z{|}_2^5=x{e}^y(5-2)=3x{e}^y$$

Now we just have to do the double integral of this over the rectangle $R=\{1\le x\le 2,0\le y\}$ containing all of the slices: $${\iiint }_{E}x{e}^{y}dV={\iint }_{R}3x{e}^{y}dA$$ Again we choose a direction to slice: starting with $dy$, we decompose this into an integrated integral as $${\iint }_{R}ex{e}^{y}dA={\int }_1^2{\int }_0^{1}3x{e}^{y}dydx$$ $$={\int }_1^{2}3x\left({\int }_0^1{e}^{y}dy\right)dx={\int }_1^{2}3x\left(e^y{|}_0^1\right)dx$$ $$={\int }_1^{2}3x(e-1)dx=\frac{3}{2}{x}^2(e-1){|}_1^2=\frac{3}{2}(4-1)(e-1)$$

15.1.2 Variables in One Bound

If the domain $E$ is described so that its bounds in at two of the variables are constants, and the third set of bounds are variables, then there is a preferred order in which to integrate. In particular, we know the final answer must be a number so we cannot have variables in the outermost set of bounds, and must be done earlier: the easiest situation is just to do it first!

Example 15.2 For example, consider the following domain $E$:

$$E=\{(x,y,z)\mid 0\le x\le 2,0\le y\le 3,0\le z\le x+y\}$$

Here the $z$ bound is different depending on which point $(x,y)$ you are at, so we do the $z$-integral first.

$$\mathrm{Slice}={\int }_0^{x+y}fdz$$

Since both the $x$ and $y$ bounds are constants, the remaining region in the $xy$ plane is a rectangle, and we know how to integrate over these.

$${\iiint }_{E}fdV={\iint }_R\mathrm{Slice}dA={\int }_0^2{\int }_0^3\mathrm{Slice}dydx$$

Putting this all together gives representation of the triple integral as an iterated integral:

$${\int }_0^2{\int }_0^3{\int }_0^{x+y}fdzdydx$$

We then evaluate this triple integral as three one dimensional integrals from Calculus I.

15.1.3 Variables in Two Bounds

For more complicated domains, its possible that variables will appear in two of the bounds. (Because the final answer must be a number, we know the outer bounds must be constants, so they cannot appear in all three bounds).

In such cases, the innermost integral can have bounds depending on two variables (the next two to be integrated), and the middle integral can have bounds depending on the outermost integral. This way, at each stage the function only has variables left in it that are still going to be integrated away, and the result is a number. In this case, there is only one possible order in which the integral can be performed!

Example 15.3 Here’s an example: if $E$ is the following region

$$E=\{-y\le x\le yz,0\le y\le z+1,-1\le z\le 1\}$$

Looking at the bounds, we notice the following:

• The $x$ bounds depend on both $y$ and $z$
• The $y$ bounds depend on $z$
• The $z$ bounds are constants.

This suggests a most natural order of integration: we start with the $x$ integral as our first slice, as its bounds depend on the remaining integrals $y$ and $z$. Then we do the $y$ integral, as its bounds depend on the remaining integral $z$. Finally we do the $z$ integral as its bounds are constant!

This gives us a direct way to write our integral as an iterated integral, which we can then solve by doing three integrals from Calculus I and II.

$${\iiint }_{E}fdV={\int }_{-1}^1{\int }_0^{z+1}{\int }_{-y}^{yz}fdxdydz$$

Sometimes we have to do some work to sovle for the bounds given equations describing the region $E$. Below is one such example:

Example 15.4 Write the integral of $f$ over the region $E$ as an iterated integral, where $E$ is bounded by the $xy$ plane and the surface $z=1-x^2-y^2$.

Since the $xy$ plane is given by $z=0$, the surfaces we are given directly describe the $z$ bounds, which suggests we begin with the $dz$ integral. $$\mathrm{Slice}={\int }_0^{1-x^2-y^2}fdz$$

But now we are left to discover the remaining bounds for ourselves! How do we do this? The region $R$ over which we need to add up the $z$ slices can be described as the points $(x,y)$ where the $z$ bounds define some interval of integration. The boundary of this region is where the $z$ bounds collapse to become equal! That is, $R$ is cut out by what we get from equating the bounds $$z=1-x^2-y^2\mathrm{and}z=0$$ This implies $x^2+y^2=1$, or the region $R$ is bounded by the unit circle. We are now to a double integral where we must again choose an order of slicing. If the bounds we had were given in $x=$ or $y=$ form such a choice might be obvious, but here there’s no natural one so we just need to choose, and solve for the correct bounds.

Choosing to do $y$ next, we solve to get $y=\pm \sqrt{1-x^2}$ as the upper and lower bounds of the integral. The interval for $x$ integration is similarly given by the points on the line where these $y$ bounds define an interval: so its boundary is where they collapse to be equal! Setting $\sqrt{1-x^2}=-\sqrt{1-x^2}$ we see the only solution is where this quantity is zero, so $\sqrt{1-x^2}=0$ or $x^2=1$ so $x=\pm 1$. This gives the interated integral

$${\iiint }_{E}fdV={\iint }_R\mathrm{Slices}dA={\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\mathrm{Slices}dydx$$

Finally putting it all together gives us a description as an iterated integral, which we could solve by doing three single variable integrals:

$${\iiint }_{E}fdV={\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{\int }_0^{1-x^2-y^2}fdzdydx$$

This last example looks rather intimidating with the square roots in the bounds, as this means the final integral we have to do will likely involve $\sqrt{1-x^2}$ after we plug these in. Such integrals often require trigonometric substitution to evaluate, and while this is often possible, we will learn in the next section a way to simplify such integrals with a change of coordinates.

15.2 Describing the Bounds:

The thing that makes triple integrals challenging is not doing the integrals (its just three 1D integrals) or even choosing the order to do them in (as we saw above, once you have described the domain in terms of $x,y,z$, its easy to decide which order to do the integral.) The difficult part is often just describing the bounds themselves!

This is mostly because visualizing 3D geometry takes some training to get used to! It’s helpful to look through many examples: please remember to be using the book (chapter 15), where each chapter is essentially just a giant list of example problems fully worked out! Additionally, Here is another collection of fully worked examples online:.

This video does a good job of explaining the process (and advocating you going and doing lots of your own practice!)

15.3 Video Resources