Triple Integrals over General Regions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.4: “Triple Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-4-triple-integrals |
| Textbook used in class | Stewart, Calculus, Section 15.7: “Triple Integrals” (Examples 2, 3, 4) |
Quick Reference
Type 1 solid: $E = \{(x,y,z): (x,y)\in D,\; u_1(x,y) \leq z \leq u_2(x,y)\}$. $$\iiint_E f\,dV = \iint_D\left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\right]dA.$$
The outer double integral $\iint_D\cdots\,dA$ is then evaluated using the Type I or Type II methods.
Motivation
Just as double integrals over general regions require variable limits for one of the two integrals, triple integrals over general 3D solids require variable limits. The most common setup (Type 1) fixes the $z$-range to be bounded above and below by surfaces $z = u_1(x,y)$ and $z = u_2(x,y)$, and describes the projection of the solid onto the $xy$-plane as a 2D region $D$.
Reading the limits from a sketch of the solid is the key skill: for each point $(x,y)$ in the projection $D$, what are the lowest and highest $z$-values in the solid?
Key Concepts
1. Setting Up the Iterated Integral
Sketch the solid. Identify:
- The bottom surface: $z = u_1(x,y)$.
- The top surface: $z = u_2(x,y)$.
- The projection $D$ onto the $xy$-plane.
Then write: $\iiint_E f\,dV = \iint_D\int_{u_1}^{u_2} f\,dz\,dA$, and expand $dA$ using the appropriate limits for $D$.
2. Six Possible Orders
For a general solid, there are six orders of integration ($dx\,dy\,dz$, $dx\,dz\,dy$, etc.). Each corresponds to a different choice of “slicing direction.” The goal is to find the order that makes the limits simplest or the integrand most tractable.
Worked Example
Find the volume of the solid bounded by the paraboloid $z = x^2+y^2$ and the plane $z = 4$. (Stewart 15.7, Example 2.)
The solid is the region between $z = x^2+y^2$ (below) and $z = 4$ (above).
Projection onto the $xy$-plane: $x^2+y^2 \leq 4$ (disk of radius 2).
$$V = \iint_D\int_{x^2+y^2}^4 dz\,dA = \iint_D (4 - x^2 - y^2)\,dA.$$
In polar ($D$: $0 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$): $$V = \int_0^{2\pi}\int_0^2(4-r^2)\,r\,dr\,d\theta = 2\pi\int_0^2(4r-r^3)\,dr = 2\pi\left[2r^2-\frac{r^4}{4}\right]_0^2 = 2\pi(8-4) = 8\pi.$$
the projection $D$ is always a simple shape like a rectangle or circle. The projection $D$ can be any 2D region -- a triangle, an ellipse, or a more complex shape bounded by curves. Reading the projection correctly requires identifying which $(x,y)$ pairs correspond to points actually inside the solid.
Leveled Practice
Problem 1. Set up (do not evaluate) the triple integral for the volume of the tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$.
Show answer
The bounding plane is $x+y+z = 1$, or $z = 1-x-y$.
Projection: the triangle $D$ with $x \geq 0$, $y \geq 0$, $x+y \leq 1$.
Type I: $0 \leq x \leq 1$, $0 \leq y \leq 1-x$.
$$V = \int_0^1\int_0^{1-x}\int_0^{1-x-y} dz\,dy\,dx.$$
(Evaluates to $1/6$.)