Triple Integrals


Motivated by the concept of double integrals, we want to consider the integrals of functions of three variables. The domain of integration will now be a three-dimensional solid region.

Triple integrals over rectangular boxes

Integral sums

Assume that \(f\) is a function defined on a rectangular box \([a,b]\times [c,d]\times [e,g]\) in the space \(\mathbb R^3\).

Consider three sequences of points \begin{eqnarray*}&&a=x_0\leq x_1\leq x_2\leq\cdots\leq x_n=b,\newline &&c=y_0\leq y_1\leq y_2\leq\cdots\leq y_m=d,\mbox{ and}\newline &&e=z_0\leq z_1\leq z_2\leq\cdots\leq z_p=g\end{eqnarray*} for \(m,n,p\in\mathbb N\). The box \([a,b]\times[c,d]\times[e,g]\) is the union of rectangles \([x_i,x_{i+1}]\times[y_j,y_{j+1}]\times[z_k,z_{k+1}]\) for each choice of \(i\), \(j\), and \(k\) such that \(i\in\{0,1,\dots, n-1\}\), \(j\in\{0,1,\dots, m-1\}\), and \(k\in\{0,1,\dots, p-1\}\). Denote these rectangles by \(R_1\), \(R_2\), \(\dots\), \(R_{mnp}\). Denote by \(A_i\) the area of the rectangle \(R_i\) and let \(c_i\) be an arbitrary point from the rectangle \(R_i\). The integral sum (also known as the Riemann sum) corresponding to the partition into rectangles \(R_1\), \(\dots\), \(R_{mnp}\) with the points \(c_1\), \(\dots\), \(c_{mnp}\) is defined as \[S(f,R_1,\dots, R_{mnp}, c_1,\dots, c_{mnp})= f(c_1)\cdot A_1+ f(c_2)\cdot A_2+\cdots + f(c_{mnp})\cdot A_{mnp}.\] The diameter of partition is defined as \[\delta(R_1, \dots, R_{mnp})=\max\{A_1,\dots ,A_{mnp}\}.\]

Definition of triple integrals over rectangular boxes

Consider the function \(f:[a,b]\times[c,d]\times[e,g]\to \mathbb R\), and consider the partitions of the box \([a,b]\times[c,d]\times[e,g]\) into smaller boxes. We say that the function \(f\) is integrable if the integral sums \(S(f,R_1, \dots, R_{mnp}, c_1, \dots, c_{mnp})\) converge as the diameters of partitions converge to \(0\). Here is the formal definition:

Definition: Triple integral over rectangular box The function \(f:[a,b]\times[c,d]\times[e,g]\to\mathbb R\) is integrable and its integral over the box \([a,b]\times[c,d]\times[e,g]\) is equal to \(I\in\mathbb R\) if for each \(\varepsilon > 0\) there exists \(\delta > 0\) such that for every partition \(\{R_1,\dots, R_{mnp}\}\) of \([a,b]\times[c,d]\times[e,f]\) with \(\delta(R_1,\dots, R_{mnp}) < \delta\) we have \[\left|S(f,R_1,\dots, R_{mnp},c_1,\dots, c_{mnp})-I\right| < \varepsilon\] for every choice of \(c_1\in R_1\), \(c_2\in R_2\), \(\dots\), \(c_{mnp}\in R_{mnp}\).

The value \(I\) from the previous definition is often denoted as \[I=\iiint_{[a,b]\times[c,d]\times[e,g]} f(x,y,z)\,dxdydz \quad\quad\quad \mbox{or} \quad\quad\quad I=\iiint_{[a,b]\times[c,d]\times[e,g]} f(x,y,z)\,dV.\]

Iterated integrals and Fubini’s theorem

Assume that \(f:[a,b]\times[c,d]\times[e,g]\to\mathbb R\) is an integrable function. For each fixed \((x,y)\in[a,b]\times[c,d]\) the function \(z\mapsto f(x,y,z)\) is a function of one variable and we can talk about single integral \(\int_e^g f(x,y,z)\,dz\). The result of integration will be a function in \((x,y)\) so we can talk about \(\int_a^b\int_c^d \left(\int_e^g f(x,y,z)\,dz\right)\,dydx\). This is called the iterated integral.

Fubini’s theorem in three dimension states that the iterated integrals are equal to the triple integral defined above.

Theorem (Fubini) If \(f:[a,b]\times[c,d]\times[e,g]\to\mathbb R\) is a continuous function, then \begin{eqnarray*}\iiint_{[a,b]\times[c,d]\times[e,g]} f(x,y,z)\,dxdydz&=&\int_a^b \int_c^d\int_e^g f(x,y,z)\,dzdydx \newline &=& \int_a^b\int_e^g \int_c^d f(x,y,z)\,dydzdx\newline &=& \int_c^d\int_a^b \int_e^g f(x,y,z)\,dzdxdy\newline &=& \int_c^d \int_e^g\int_a^b f(x,y,z)\,dxdzdy\newline &=& \int_e^g\int_a^b \int_c^d f(x,y,z)\,dydxdz \newline &=& \int_e^g\int_c^d\int_a^b f(x,y,z)\,dxdydz .\end{eqnarray*}

Remark. Fubini’s theorem holds also if instead of continuity of \(f\) we assumed that \(f\) is bounded on \([a,b]\times[c,d]\times[e,g]\), the set of discontinuities consists of finitely many smooth surfaces, and the iterated integrals exist.

We will omit the proof of the theorem.

Triple integrals over general regions

Our goal now is to generalize the integration to include functions that are defined on regions that are not rectangular boxes.

Assume now that \(D\subseteq \mathbb R^3\) is a general region that is not necessarily a rectangular box. We can define the integral \(\iint_D f(x,y,z)\,dV\) in the following way. First, assume that \(R\) is a rectangle of the form \([a,b]\times[c,d]\times[e,g]\) for some \(a,b,c,d,e,g\in\mathbb R\). Then we define the function \(\hat f:R\to\mathbb R\) in the following way \[\hat f(x,y,z)=\left\{\begin{array}{ll} f(x,y,z),& \mbox{ if } (x,y,z)\in D,\newline 0,&\mbox{ if } (x,y,z)\not\in D,\end{array}\right.\] and we define \(\iiint_Df(x,y,z)\,dV\) as: \[\iint_Df(x,y,z)\,dV=\iiint_R\hat f(x,y,z)\,dV.\]

Practice problems

Problem 1. Evaluate the integral \[\iiint_S xyz\,dxdydz,\] where \(S\) is the rectangular box \(S=[0,2]\times[1,5]\times[-1,3]\).

Problem 2. Evaluate the integral \[\iiint_S z\,dxdydz,\] where \(S\) is a triangular prism bounded by the planes \(x=0\), \(y=0\), \(z=0\), \(x=5\), and \(y+z=1\).

Problem 3. Find the volume of the solid in the first octant that is located between the surface \(y=z^2\) and the plane \(x+y=4\).

Problem 4. Evaluate the integral \[\int_{-1}^1\int_0^{z^2}\int_0^y \frac{z+\tan\frac{z}{x^2+y^2+1}}{x^2+y^2+1}\,dxdydz.\]

Problem 5. For any bounded continuous function \(f\) the integral \(\int_0^x\int_0^y\int_0^zf(t)\,dtdzdy\) is equal to:
  • (A) \(0\)
  • (B) \(\int_0^xxf(t)\,dt\)
  • (C) \(\frac12\int_0^x(x-t)^2f(t)\,dt\)
  • (D) \(\int_0^x(y-x)f(y)dy\)
  • (E) \(xf(x)-\frac{x^2}2\)