
# 3.3: Triple Integrals

Our definition of a double integral of a real-valued function $$f (x, y)$$ over a region $$R$$ in $$\mathbb{R}^2$$ can be extended to define a triple integral of a real-valued function $$f (x, y, z)$$ over a solid $$S$$ in $$\mathbb{R}^ 3$$ . We simply proceed as before: the solid $$S$$ can be enclosed in some rectangular parallelepiped, which is then divided into subparallelepipeds. In each subparallelepiped inside $$S$$, with sides of lengths $$\Delta x, \Delta y \text{ and }\Delta z$$, pick a point $$(x∗, y∗, z∗)$$. Then define the triple integral of $$f (x, y, z)$$ over $$S$$, denoted by $$\iiint\limits_S f (x, y, z)dV$$, by

$\iiint\limits_S f (x, y, z)dV=\lim \sum\sum\sum f (x∗, y∗, z∗)\Delta x \Delta y \Delta z \label{Eq3.7}$

where the limit is over all divisions of the rectangular parallelepiped enclosing $$S$$ into subparallelepipeds whose largest diagonal is going to 0, and the triple summation is over all the subparallelepipeds inside $$S$$. It can be shown that this limit does not depend on the choice of the rectangular parallelepiped enclosing $$S$$. The symbol $$dV$$ is often called the volume element.

Physically, what does the triple integral represent? We saw that a double integral could be thought of as the volume under a two-dimensional surface. It turns out that the triple integral simply generalizes this idea: it can be thought of as representing the hypervolume under a three-dimensional hypersurface $$w = f (x, y, z)$$ whose graph lies in $$\mathbb{R}^ 4$$ . In general, the word “volume” is often used as a general term to signify the same concept for any $$n$$-dimensional object (e.g. length in $$\mathbb{R}^1$$ , area in $$\mathbb{R}^2$$ ). It may be hard to get a grasp on the concept of the “volume” of a four-dimensional object, but at least we now know how to calculate that volume!

In the case where $$S$$ is a rectangular parallelepiped $$[x_1 , x_2] \times [y_1 , y_2] \times [z_1 , z_2]$$, that is, $$S = {(x, y, z) : x_1 ≤ x ≤ x_2 , y_1 ≤ y ≤ y_2 , z_1 ≤ z ≤ z_2}$$, the triple integral is a sequence of three iterated integrals, namely

$\iiint\limits_S f (x, y, z)dV = \int_{z_1}^{z_2} \int_{y_1}^{y_2} \int_{x_1}^{x_2} f (x, y, z)\,dx\, d y\, dz \label{Eq3.8}$

where the order of integration does not matter. This is the simplest case.

A more complicated case is where $$S$$ is a solid which is bounded below by a surface $$z = g_1(x, y)$$, bounded above by a surface $$z = g_2(x, y)$$, $$y$$ is bounded between two curves $$h_1(x) \text{ and }h_2(x)$$, and $$x$$ varies between $$a \text{ and }b$$. Then

$\iiint\limits_S f (x, y, z)dV = \int_a^b \int_{h_1(x)}^{h_2(x)} \int_{g_1(x,y)}^{g_2(x,y)}f (x, y, z)\,dz\, d y\, dx \label{Eq3.9}$

Notice in this case that the first iterated integral will result in a function of $$x$$ and $$y$$ (since its limits of integration are functions of $$x \text{ and }y$$), which then leaves you with a double integral of a type that we learned how to evaluate in Section 3.2. There are, of course, many variations on this case (for example, changing the roles of the variables $$x, y, z$$), so as you can probably tell, triple integrals can be quite tricky. At this point, just learning how to evaluate a triple integral, regardless of what it represents, is the most important thing. We will see some other ways in which triple integrals are used later in the text.

Example 3.7

Evaluate $$\int_0^3 \int_0^2 \int_0^1 (x y+ z)\,dx\, d y\, dz$$.

Solution

\nonumber \begin{align} \int_0^3 \int_0^2 \int_0^1 (x y+ z)\,dx\, d y\, dz &= \int_0^3 \int_0^2 \left ( \dfrac{1}{2}x^2y+xz \big |_{x=0}^{x=1} \right )\,dy\,dz \\ \nonumber &=\int_0^3 \int_0^2 \left ( \dfrac{1}{2}y+z \right )\,dy\,dz \\ \nonumber &= \int_0^3 \left ( \dfrac{1}{4}y^2+yz \big |_{y=0}^{y=2} \right )\,dz \\ \nonumber &= \int_0^3 (1+2z)\,dz \\ \nonumber &=z+z^2 \big |_0^3 = 12 \end{align}

Example 3.8

Evaluate $$\int_0^1 \int_0^{1-x} \int_0^{2-x-y} (x+ y+ z)\,dz\, d y\, dx$$.

Solution

\nonumber \begin{align} \int_0^1 \int_0^{1-x} \int_0^{2-x-y}(x+ y+ z)\,dz\, d y\, dx &= \int_0^1 \int_0^{1-x}\left ((x+ y)z + \dfrac{1}{2}z^2 \big |_{z=0}^{z=2-x-y} \right )\,dy\,dx \\ \nonumber &=\int_0^1 \int_0^{1-x} \left ( (x+ y)(2− x− y)+ \dfrac{1}{2} (2− x− y)^2 \right )\,dy\,dx \\ \nonumber &= \int_0^1 \int_0^{1-x}\left ( 2− \dfrac{1}{2}x^2 − x y− \dfrac{1}{2}y^2 \right ) \, dy\,dx \\ \nonumber &= \int_0^1 \left ( 2y− \dfrac{1}{2} x^2 y− x y− \dfrac{1}{2} x y^2 − \dfrac{1}{6}y^3 \big |_{y=0}^{y=1-x} \right )\,dx \\ \nonumber &=\int_0^1 \left ( \dfrac{11}{6}-2x+\dfrac{1}{6}x^3 \right )\,dx \\ \nonumber &= \dfrac{11}{6}x-x^2+\dfrac{1}{24}x^4 \big |_0^1 = \dfrac{7}{8} \end{align}

Note that the volume $$V$$ of a solid in $$\mathbb{R}^ 3$$  is given by

$V= \iiint\limits_S 1dV \label{Eq3.10}$

Since the function being integrated is the constant 1, then the above triple integral reduces to a double integral of the types that we considered in the previous section if the solid is bounded above by some surface $$z = f (x, y)$$ and bounded below by the $$x y$$-plane $$z = 0$$. There are many other possibilities. For example, the solid could be bounded below and above by surfaces $$z = g_1(x, y)\text{ and }z = g_2(x, y)$$, respectively, with $$y$$ bounded between two curves $$h_1(x)\text{ and }h_2(x)$$, and $$x$$ varies between $$a \text{ and }b$$. Then

$\nonumber V=\iiint\limits_S 1dV = \int_a^b \int_{h_1(x)}^{h_2(x)} \int_{g_1(x,y)}^{g_2(x,y)} 1\,dz\, d y\, dx = \int_a^b \int_{h_1(x)}^{h_2(x)} (g_2(x, y)− g_1(x, y)) \,d y\, dx$

just like in Equation \ref{Eq3.9}. See Exercise 10 for an example.