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Math 344: Calculus III

15.6 Triple Integrals

15.6.1 Definition of a Triple Integral ¶

Integrating with respect to three dimensions is nearly the same process as integrating with respect to two dimensions, just with an extra variable of integration. The simplest example is integration of a 3D function over a rectangular prism or box of the form

$$ R = \left\{(x,y,z)\ :\ x\in[a,b],\, y\in[c,d],\, z\in[e,f]\right\} $$
This box is then subdivided into smaller boxes and transformed into a triple Riemann sum .

On the left a rectangular prism is shown divided into small boxes, with each box representing a differential element of the 3D region. On the right, the outline of the prism is shown with the region empty save for a single differential element.

The outline of a cube labeled above as R_ijk is depicted in 3D, with dimension markings of Delta x, Delta y, and Delta z representing the length, width, and height of the cube. Inside the cube, a sample point is labeled, indicating a random selection of where to evaluate f(x,y,z).

The region $R$ is divided into $lmn$ subintervals where $l$, $m$, and $n$ are the total number of subintervals for $x$, $y$, and $z$ and each differential element is defined by a small change in the $x$, $y$, and $z$ values, $\Delta x$, $\Delta y$, and $\Delta z$, just as with the Riemann sums we have seen before. Inside of this small box $R_{ijk}$, a sample point $(x^*_{ijk},y^*_{ijk},z^*_{ijk})$ is chosen to evaluate the function $f(x,y,z)$ in order to obtain an approximation to the value of that function in the differential element. This is done for each element and the limit of the this sum is taken as the size of the elements go to zero. The differential volume is indicated in the integral as $dV$.

Definition ¶

The triple integral of a function $f(x,y,z)$ over a 3D region $R$ is given by

$$ \iiint_R f(x,y,z)\,dV = \lim_{l,\,m,\,n\rightarrow\infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(x_{ijk}^*,y_{ijk}^*,z_{ijk}^*)\,\Delta V $$
where $i$, $j$, and $k$ are indices indicating indicating the differential element $R_{ijk}$.

15.6.2 Fubini's Theorem ¶

There is no significant change to handling triple integrals when compared to double integrals. We set them up as iterated integrals and compute with respect to one dimension at a time.

Fubini's Theorem for Triple Integrals ¶

If $f$ is continuous on the rectangular prism (box) $R = [a,b]\times [c,d]\times [r,s]$, then

$$ \iiint_R f(x,y,z)\,dV = \int_r^s \int_c^d \int_a^b f(x,y,z)\,dxdydz $$

As in the 2D case, if the integration domain is a rectangular prism we can switch the order of integration. This means that there are six possible orders of integration, and they all yield the same value. Two possible alternative orders are

$$ \iiint_R f(x,y,z)\,dV = \int_c^d \int_r^s \int_a^b f(x,y,z)\,dxdzdy = \int_r^s \int_a^b \int_c^d f(x,y,z)\,dydxdz $$

Example 1 ¶

Evaluate the integral $\iiint_E \left( xy^2 + z\right)dV$ where

$$ E = \left\{ (x,y,z)\ :\ x\in[0,2],\, y\in[0,1],\, z\in[-1,1]\right\} $$
using two different orders of integration.

Solution ¶

Order 1

$$ \begin{align*} \int_0^1 \int_{-1}^1 \int_0^2 \left( xy^2 + z\right)\,dxdzdy &= \int_0^1 \int_{-1}^1 \left.\left[ \frac{1}{2}x^2y^2 + xz\right]\right|_{\,0}^{\,2}\,dzdy \\ \\ &= \int_0^1 \int_{-1}^1 2y^2 + 2z\,dzdy \\ \\ &= \int_0^1 \left.\left[ 2y^2 z + z^2 \right]\right|_{\,-1}^{\,1}\,dy \\ \\ &= \int_0^1 \left(2y^2 + 1^2\right) - \left(-2y^2 + (-1)^2\right)\,dy \\ \\ &= \int_0^1 4y^2\,dy \\ \\ &= \left. \frac{4}{3}y^3\, \right|_{\,0}^{\,1} \\ \\ &= \frac{4}{3} \end{align*} $$

Order 2

$$ \begin{align*} \int_{-1}^1 \int_0^2 \int_0^1 \left( xy^2 + z\right)\,dydxdz &= \int_{-1}^1 \int_0^2 \left.\left[ \frac{1}{3}xy^3 + yz\right]\right|_{\,0}^{\,1}\,dxdz \\ \\ &= \int_{-1}^1 \int_0^2 \frac{1}{3}x + z\,dxdz \\ \\ &= \int_{-1}^1 \left.\left[ \frac{1}{6}x^2 + xz \right]\right|_{\,0}^{\,2}\,dz \\ \\ &= \int_{-1}^1 \frac{2}{3} + 2z\,dz \\ \\ &= \left. \frac{2}{3}z + z^2\, \right|_{\,-1}^{\,1} \\ \\ &= \left(\frac{2}{3} + 1^2\right) - \left(-\frac{2}{3} + (-1)^2\right) \\ \\ &= \frac{4}{3} \end{align*} $$

Choose a third order to try on your own.

15.6.3 Integrating Over General Regions ¶

Constructing iterated integrals for general 3D regions follows the same procedure as in the 2D case. If we wish to find the value of $ \iiint_R f(x,y,z)\,dV $, the first step is to find a rectangular prism containing $R$ and defining a new function $F(x,y,z)$ on that prism with the property

$$ F(x,y,z) = \left\{\begin{array}{cl} f(x,y,z) & (x,y,z)\in R \\ 0 & \text{otherwise} \end{array}\right. $$

Under this condition, Fubini's theorem will apply to the triple integral and we seek functions that bound the shape in order to construct an iterated integral. Once these functions are determined, we set up and evaluate the integral. An example of one such problem is depicted in the figure below.

On a 3D coordinate system, a shape D with curved edges on the left and right side is plotted in the xy-plane. Suspended above the shape is a 3D region R bounded on the left and right and front and back by the lower 2D shape. It is bounded above by functions u1(x,y) and u2(x,y), with several labels indicating the boundaries of the 3D region.

Here we have a region where the bottom and top boundaries may be expressed as functions of $x$ and $y$, called $u_1(x,y)$ and $u_2(x,y)$ respectively. On the left and right, we see that the region is bounded by functions of $x$, $g_1(x)$ and $g_2(x)$. Finally, the front and back boundaries given by the interval $[a,b]$. This suggests the following iterated integral:

$$ \iiint_R f(x,y,z)\,dV = \int_a^b \int_{y\,=\,g_1(x)}^{y\,=\,g_2(x)} \int_{z\, =\, u_1(x,y)}^{z\, =\, u_2(x,y)} f(x,y,z)\,dzdydx $$
The basic structure of this integral is

$$ \iiint_R f(x,y,z)\,dV = \iint_D \left[ \int_{\, u_1(x,y)}^{\, u_2(x,y)} f(x,y,z)\,dz \right]dA $$
where after the computation of the interior integral, you are left with a standard double integral.

Example 2 ¶

Evaluate the iterated integral

$$ \int_0^1 \int_0^1 \int_0^{2-x^2-z^2} xze^y\,dydzdx $$

Solution ¶

$$ \begin{align*} \int_0^1 \int_0^1 \int_0^{2-x^2-z^2} xze^y\,dydzdx &= \int_0^1 \int_0^1 \left.\left[ xze^y \right]\right._{\,0}^{\,2-x^2-z^2}\,dzdx \\ \\ &= \int_0^1 \int_0^1 \left( xze^{2-x^2-z^2} - xz\right)\,dzdx \\ \\ &= \int_0^1 \left.\left[ -\frac{1}{2}xe^{2-x^2-z^2} - \frac{1}{2}xz^2\right]\right|_{\,0}^{\,1}\,dx \\ \\ &= \frac{1}{2} \int_0^1 -xe^{1-x^2} - x + xe^{2-x^2}\,dx \\ \\ &= \frac{1}{2} \left.\left[ \frac{1}{2}e^{1-x^2} - \frac{1}{2}x^2 - \frac{1}{2}e^{2-x^2}\right]\right|_{\,0}^{\,1} \\ \\ &= \frac{1}{4}\left(1 - 1 - e - e + 0 + e^2\right) \\ \\ &= \frac{1}{4}\left(e^2 - 2e\right) \end{align*} $$

The solid region bounded on the left by y = 0 and on the right by y = 2 - x^2 - z^2 for x in [0,1] and z in [0,1] is depicted on a 3D coordinate plane.

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