Triple Integrals over a General Region | Tony's Intertwingled Memex

Core Definition

A triple integral over a general 3D bounded region $E \subset \mathbb{R}^3$ is defined by enclosing $E$ within a larger rectangular box $B$ and setting an indicator integrand $F(x, y, z) = f(x, y, z)$ on $E$ and $0$ outside $E$ :

\iiint_{E} f(x, y, z) \, dV = \iiint_{B} F(x, y, z) \, dV

For a Type I region, the solid is bounded between continuous surfaces

z = u_1(x, y)

and

z = u_2(x, y)

over a planar 2D projection domain

D \subset \mathbb{R}^2

in the

xy

-plane:

\iiint_{E} f(x, y, z) \, dV = \iint_{D} \left( \int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z) \, dz \right) dA

Concept Explanation

In real physical and engineering problems, 3D solids are rarely simple rectangular boxes. Instead, boundaries are curved surfaces, slanted planes, and intersecting manifolds. We evaluate triple integrals over such general regions by setting functional limits on the inner variables while maintaining constant limits on the outermost variable.

Region Classifications in 3D Space

Depending on which coordinate axis connects the bottom and top bounding surfaces, 3D regions are classified into three fundamental types:

Type I ( $z$ -simple): $E = \{(x, y, z) \mid (x, y) \in D_{xy}, \, u_1(x, y) \le z \le u_2(x, y)\}$ The solid is bounded below by $z = u_1(x, y)$ and above by $z = u_2(x, y)$ . The 2D shadow $D_{xy}$ lies in the $xy$ -plane.
Type II ( $x$ -simple): $E = \{(x, y, z) \mid (y, z) \in D_{yz}, \, u_1(y, z) \le x \le u_2(y, z)\}$ The solid is bounded behind by $x = u_1(y, z)$ and in front by $x = u_2(y, z)$ . The 2D shadow $D_{yz}$ lies in the $yz$ -plane.
Type III ( $y$ -simple): $E = \{(x, y, z) \mid (x, z) \in D_{xz}, \, u_1(x, z) \le y \le u_2(x, z)\}$ The solid is bounded on the left by $y = u_1(x, z)$ and on the right by $y = u_2(x, z)$ . The 2D shadow $D_{xz}$ lies in the $xz$ -plane.

The 4-Step Boundary Construction Protocol

To set up an iterated triple integral over a general region (e.g., standard order $dz \, dy \, dx$ ):

Inner Surface Limits ( $z$ ): Look along the chosen inner axis ( $z$ ). Identify the entering surface $z = u_1(x, y)$ (bottom) and exiting surface $z = u_2(x, y)$ (top).
2D Projection Shadow ( $D$ ): Project the outermost boundary of the solid onto the remaining coordinate plane ( $xy$ -plane). This boundary is found by intersecting the upper and lower surfaces ( $u_1(x, y) = u_2(x, y)$ ) or setting bounding planes to coordinate axes ( $z = 0$ ).
Middle Curve Limits ( $y$ ): Treat the 2D projection $D$ as a double integral domain. Cast an arrow along the middle axis ( $y$ ) from the lower boundary curve $y = g_1(x)$ to the upper boundary curve $y = g_2(x)$ .
Outer Constant Limits ( $x$ ): Identify the absolute minimum and maximum constant extent of the shadow along the outer axis: $a \le x \le b$ .

Putting these together yields the complete iterated integral:

\iiint_{E} f(x, y, z) \, dV = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} \int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z) \, dz \, dy \, dx

Physical Quantities: Volume, Mass, and Center of Mass

Spatial Volume: $V(E) = \iiint_E 1 \, dV$ .
Total Mass: If $\rho(x, y, z)$ is mass density, then $M = \iiint_E \rho(x, y, z) \, dV$ .
First Moments: $M_{yz} = \iiint_E x \rho \, dV, \quad M_{xz} = \iiint_E y \rho \, dV, \quad M_{xy} = \iiint_E z \rho \, dV$
Center of Mass Coordinates: $\bar{x} = \frac{M_{yz}}{M}, \quad \bar{y} = \frac{M_{xz}}{M}, \quad \bar{z} = \frac{M_{xy}}{M}$

Engineering Applications

Aerospace & Mechanical Engineering: Calculating the total fuel mass and dynamic center of gravity $\vec{r}_{\text{CG}} = \langle \bar{x}, \bar{y}, \bar{z} \rangle$ for non-rectangular wingtip fuel cells, fuselage tanks, and liquid propellant slosh baffles.
Civil & Structural Engineering: Computing total concrete volume, dead load, and foundation bearing moments for asymmetric bridge abutments, gravity retaining walls, and dam spillways.
Ocean & Naval Architecture: Determining hydrostatic displacement volume and center of buoyancy for ship hulls and submarine ballast tanks bounded by parabolic waterlines and inclined transverse bulkheads.

Example Problems

Example 1: Integrating over a First-Octant Tetrahedron

Evaluate $\iiint_{E} z \, dV$ , where $E$ is the solid tetrahedron in the first octant bounded by the planes $x = 0$ , $y = 0$ , $z = 0$ , and $2x + y + z = 2$ .

Solution

Step 1: Determine inner $z$ -limits:
The solid lies above $z = 0$ and below the inclined plane $2x + y + z = 2 \implies z = 2 - 2x - y$ .
Thus, $0 \le z \le 2 - 2x - y$ .

Step 2: Determine 2D projection $D$ onto the $xy$ -plane:
Setting $z = 0$ in the plane equation gives $2x + y = 2 \implies y = 2 - 2x$ . In the first octant, $x \ge 0, y \ge 0$ .
Middle limits: $0 \le y \le 2 - 2x$ .
Outer limits: Setting $y = 0$ yields $2x = 2 \implies 0 \le x \le 1$ .

Step 3: Set up and evaluate the iterated integral:

I = \int_{0}^{1} \int_{0}^{2-2x} \int_{0}^{2-2x-y} z \, dz \, dy \, dx

Inner $z$ -integral:

\int_{0}^{2-2x-y} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{2-2x-y} = \frac{1}{2}(2 - 2x - y)^2

Middle $y$ -integral:
Use $u$ -substitution: let $u = 2 - 2x - y \implies dy = -du$ .
When $y = 0, u = 2 - 2x$ ; when $y = 2 - 2x, u = 0$ .

\int_{2-2x}^{0} \frac{1}{2}u^2 (-du) = \int_{0}^{2-2x} \frac{1}{2}u^2 \, du = \left[ \frac{u^3}{6} \right]_{0}^{2-2x} = \frac{(2 - 2x)^3}{6} = \frac{8(1 - x)^3}{6} = \frac{4}{3}(1 - x)^3

Outer $x$ -integral:

\int_{0}^{1} \frac{4}{3}(1 - x)^3 \, dx = \left[ -\frac{1}{3} \cdot \frac{4}{3}(1 - x)^4 \right]_{0}^{1} = \left[ -\frac{4}{9}(1 - x)^4 \right]_{0}^{1} = 0 - \left(-\frac{4}{9}\right) = \frac{4}{9}

Thus, $\iiint_E z \, dV = \frac{4}{9}$ .

Example 2: Volume Bounded by a Parabolic Cylinder

Find the volume of the solid $E$ bounded by the parabolic cylinder $y = x^2$ and the planes $z = 0$ and $y + z = 4$ .

Solution

Step 1: Inner $z$ -limits:
The solid extends from the floor $z = 0$ up to the ceiling $y + z = 4 \implies z = 4 - y$ .
Thus, $0 \le z \le 4 - y$ .

Step 2: 2D projection $D$ onto the $xy$ -plane:
Setting $z = 0$ in $y + z = 4$ gives $y = 4$ . The lateral boundary is $y = x^2$ .
Middle limits ( $y$ ): for each $x$ , $y$ goes from the parabola up to the line: $x^2 \le y \le 4$ .
Outer limits ( $x$ ): the intersection points are $x^2 = 4 \implies -2 \le x \le 2$ .

Step 3: Evaluate volume integral:

V = \int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} 1 \, dz \, dy \, dx = \int_{-2}^{2} \int_{x^2}^{4} (4 - y) \, dy \, dx

Middle $y$ -integral:

\int_{x^2}^{4} (4 - y) \, dy = \left[ 4y - \frac{y^2}{2} \right]_{x^2}^{4} = \left(16 - 8\right) - \left(4x^2 - \frac{x^4}{2}\right) = 8 - 4x^2 + \frac{x^4}{2}

Outer $x$ -integral (exploiting symmetry of even integrand):

V = 2 \int_{0}^{2} \left(8 - 4x^2 + \frac{x^4}{2}\right) \, dx = 2 \left[ 8x - \frac{4}{3}x^3 + \frac{x^5}{10} \right]_{0}^{2}

V = 2 \left( 16 - \frac{32}{3} + \frac{32}{10} \right) = 2 \left( 16 - \frac{32}{3} + \frac{16}{5} \right) = 2 \left( \frac{240 - 160 + 48}{15} \right) = 2\left(\frac{128}{15}\right) = \frac{256}{15}

Connections & References

Parent Concepts: Multiple Integrals, Double Integrals over a General Region
Sub-concepts & Extensions:
Course Links:
- Calc-3 Session 16 - Triple Integrals