Chapter 13 Triple Integrals

13.1 Definition of Triple Integral

Consider the function of three variables $f(x,y,z)$, and a three-dimensional region $V \subseteq \mathbb{R}^{3}$. The integral of $f$ over $V$, denoted $\int \int \int_{V} f(x,y,z) \,dx \,dy \,dz$, is known as a triple integral. The definition of the triple integral is similar to that of a double integral.

Let $\Delta x$, $\Delta y$ and $\Delta z$ be three sufficiently small real values. Consider a cuboid in the $(x,y,z)$-space of length $\Delta x$, width $\Delta y$ and height $\Delta z$. Fill $V$ as efficiently as possible with these small cuboids. A typical cuboid has volume $\Delta x \Delta y \Delta z$ and vertices: $$\begin{align*} (x_0,y_0,z_0), \qquad & \qquad (x_0+ \Delta x,y_0,z_0), \\ (x_0,y_0+ \Delta y,z_0), \qquad & \qquad (x_0,y_0,z_0+ \Delta z), \\ (x_0 + \Delta x,y_0 + \Delta y,z_0), \quad & \quad (x_0 + \Delta x,y_0,z_0 + \Delta z), \\ (x_0,y_0 + \Delta y,z_0 + \Delta z), \quad & \quad (x_0 + \Delta x,y_0 + \Delta y,z_0+ \Delta z). \end{align*}$$

One can form the sum: $$\sum\limits_{\text{cuboids}} f(x_0,y_0,z_0) \Delta x \Delta y \Delta z.$$ Consider the limiting process as $\Delta x, \Delta y, \Delta z \rightarrow 0$. As this limits, the volume of each cuboid gets smaller, and $V$ becomes more tightly packed with a greater number of cuboids. If the above summation tends to a limit, then this limit is defined as the tiple integral. Specifically

$$\sum\limits_{\text{cuboids}} f(x_0,y_0,z_0) \Delta x \Delta y \Delta z \quad \xrightarrow{\Delta x, \Delta y, \Delta z \rightarrow 0} \quad \int \int \int_V f(x,y,z) \,dx \,dy \,dz.$$

Computation of triple integrals can be difficult directly from the definition.

A triple integral can be evaluated as three nested single integrals. Specifically $$\int \int \int_V f(x,y,z) \,dx \,dy \,dz = \int_{z_1}^{z_2} \int_{y_1(z)}^{y_2(z)} \int_{x_1(y,z)}^{x_2(y,z)} f(x,y,z) \,dx \,dy \,dz,$$ where

• $z_1$ is the smallest value that the $z$-coordinate takes over $V$;

• $z_2$ is the largest value that the $z$-coordinate takes over $V$;

• $y_1(\tilde{z})$ is the smallest value of the $y$-coordinate among points in $V$ with $z$-coordinate equal to $\tilde{z}$;

• $y_2(\tilde{z})$ is the largest value of the $y$-coordinate among points in $V$ with $z$-coordinate equal to $\tilde{z}$;

• $x_1(\tilde{y},\tilde{z})$ is the smallest value that the $x$-coordinate takes among all the points in $V$ with $y$-coordinate equal to $\tilde{y}$ and $z$-coordinate equal to $\tilde{z}$;

• $x_2(\tilde{y},\tilde{z})$ is the largest value that the $x$-coordinate takes among all the points in $V$ with $y$-coordinate equal to $\tilde{y}$ and $z$-coordinate equal to $\tilde{z}$.

Calculate the integral of $f(x,y,z) = x+y z^2$ over the cube:

$$C = \left\{ (x,y,z) : 0 \leq x \leq 1, -1 \leq y \leq 2, 2 \leq z \leq 3 \right\} \subset \mathbb{R}^3.$$

By applying Lemma 13.1.1, the integral can be evaluated as:

$$\begin{align*} \int \int \int f(x,y,z) \,dx \,dy \,dz &= \int_{2}^{3} \int_{-1}^{2} \int_{0}^{1} x+y z^2 \,dx \,dy \,dz \\ &= \int_{2}^{3} \int_{-1}^{2} \left[ \frac{1}{2} x^2 + xyz^2 \right]_{0}^{1} \,dy \,dz \\ &= \int_{2}^{3} \int_{-1}^{2} \frac{1}{2} + y z^2 \,dy \,dz \\ &= \int_{2}^{3} \left[ \frac{1}{2}y + \frac{1}{2} y^2 z^2 \right]_{-1}^{2} \,dz \\ &= \int_{2}^{3} \frac{3}{2} + \frac{3}{2}z^2 \,dz \\ &= \left[ \frac{3}{2} z + \frac{1}{2} z^3 \right]_{2}^{3} \\ &= 11. \end{align*}$$

There are a number of applications of triple integrals such as center of masses, moments of inertia for 3d bodies and modelling fluid flows. The modules MATH2005: Vector Calculus and MATH2012: Modelling with Differential Equations are excellent opportunities to study more of these concepts.

Consider a three dimensional object $V$ whose density is described by $\rho(x,y,z)$. Then $$\text{Mass}(V) = \int \int \int_{V} \rho(x,y,z) \,dx \,dy \,dz.$$

Consider a three dimensional object $V$. Define $$\bar{x} = \frac{\int \int \int_V x \,dx \,dy\,dz}{\int \int \int_V 1 \,dx \,dy\,dz},$$ and similarly define $\bar{y}$ and $\bar{z}$. Then the center of mass of $V$ is given by $(\bar{x},\bar{y},\bar{z})$.

13.2 Change of Variables for Triple Integral

Consider a one-to-one transformation $x=x(u,v,w)$, $y=y(u,v,w)$ and $z=z(u,v,w)$. Suppose one wanted to rewrite a triple integral $\int \int \int_{V} f(x,y,z) \,dx \,dy \,dz$ in terms of the new variables $u,v,w$. There is a result analogous to Theorem 12.2.3 that allows one to do this.

$$\int \int \int_V f(x,y,z) \,dx \,dy \,dz = \int \int \int_{V'} g(u,v,w) \lvert J \rvert \,du \,dv \,dw,$$ where

• $g(u,v,w) = f \big( x(u,v,w), y(u,v,w), z(u,v,w) \big)$;

• $V' = \left\{ (u_0,v_0,w_0) : \big( x_0(u_0,v_0,w_0), y_0(u_0,v_0,w_0), z_0(u_0,v_0,w_0) \big) \in V \right\}$;

• $J$ is the Jacobian of the transformation $$J = \frac{\partial(x,y,z)}{\partial (u,v,w)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix}.$$

The Jacobian $J$ for a one-to-one transformation of three variables maintains many of the properties seen for the Jacobian of a two-variable transformation. In particular, $J$ describes the volume in $(x,y,z)$-space associated with small coordinate changes $\Delta u,\Delta v, \Delta w$ in $(u,v,w)$-space. Also provided a Jacobian is non-zero, then the Jacobian of the inverse is the reciprical: $\frac{\partial (u,v,w)}{\partial (x,y,z)} = \frac{1}{\left( \frac{\partial (x,y,z)}{\partial (u,v,w)} \right)}$.

Evaluate $I = \int \int \int_D x^2 + y^2 \,dx \,dy \,dz$ where $D$ is the region in the first octant ($x,y,z \geq 0$) bounded by the cylinders $x^2 + y^2 =1$ and $x^2 + y^2 =4$, and the planes $z=0$, $z=1$, $x=0$ and $x=y$.

Due to the cylindrical nature of the problem, it may be beneficial to switch to cylindrical coordinates. In particular $$\begin{align*} x &= \rho \cos (\varphi), \\ y &= \rho \sin (\varphi), \\ z &= z. \end{align*}$$

Note $f(x,y,z) = x^2 + y^2$, so

$$\begin{align*} g(\rho, \varphi, z) &= \big( \rho \cos(\varphi) \big)^2 + \big( \rho \sin(\varphi) \big)^2 \\ &= \rho^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big) \\ &= \rho^2. \end{align*}$$

Now looking at the bounding lines of $D$, and rewriting them in terms of $\rho, \varphi$ and $z$ by substitution, one obtains $$\begin{align*} x^2 + y^2 =1 \qquad &\text{becomes} \qquad \rho=1, \\[3pt] x^2 + y^2 =4 \qquad &\text{becomes} \qquad \rho=2, \\[3pt] x=0 \qquad &\text{becomes} \qquad \varphi=\frac{\pi}{2}, \\[3pt] y=x \qquad &\text{becomes} \qquad \varphi=\frac{\pi}{4}, \\[3pt] z=0 \qquad &\text{becomes} \qquad z=0 \\[3pt] z=1 \qquad &\text{becomes} \qquad z=1. \end{align*}$$ Therefore $D$ can be written as $$D = \left\{ (\rho, \varphi,z): 1\leq \rho \leq 2, \frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2}, 0 \leq z \leq 1 \right\}.$$

Calculating the Jacobian, one obtains

$$\begin{align*} J &= \frac{\partial (x,y,z)}{\partial (\rho, \varphi,z)} \\[5pt] &= \begin{vmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \varphi} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \varphi} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \varphi} & \frac{\partial z}{\partial z} \end{vmatrix} \\[5pt] &= \begin{vmatrix} \cos \varphi & - \rho \sin \varphi & 0 \\ \sin \varphi & \rho \cos \varphi & 0 \\ 0 & 0 & 1 \end{vmatrix} \\[5pt] &= \rho. \end{align*}$$

Applying Theorem 13.2.1 one has

$$\begin{align*} I &= \int_{1}^{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{1} \rho^2 \cdot \lvert \rho \rvert \,dz \,d\varphi \,\rho \\[3pt] &= \int_{1}^{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{1} \rho^3 \,dz \,d\varphi \,\rho \\[3pt] &= \int_{1}^{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \rho^3 \,d\varphi \,\rho \\[3pt] &= \int_{1}^{2} \frac{\pi}{4} \rho^3 \,\rho \\[3pt] &=\left[ \frac{\pi}{4} \cdot \frac{\rho^4}{4}\right]_{1}^{2} \\[3pt] &= \frac{15 \pi}{16}. \end{align*}$$