Triple Integrals and Solid Bounds

Build a solid from slices

A triple integral $\iiint_E f\,dV$ adds small contributions $f\Delta V$ . If $f=1$ , it gives volume. If $f$ is mass density, it gives mass. For a solid between $z=g_1(x,y)$ and $z=g_2(x,y)$ above a planar region $R$ , integrate in $z$ first:

\iiint_Ef\,dV=\iint_R\int_{g_1(x,y)}^{g_2(x,y)}f(x,y,z)dz\,dA.

Projection onto the outer coordinate plane determines $R$ . Bound expressions may depend only on variables not yet integrated; otherwise the iterated integral is not properly specified.

Visual guide

VISUAL GUIDEThe tetrahedron’s roof sets the inner bound

The projected solid x, y, z ≥ 0, x + y + z ≤ 1 has intercepts at the three unit axes. At each point in its triangular xy footprint, z runs from 0 to 1 − x − y. This projection illustrates the bounds, not true lengths.

Worked example: a tetrahedron

The solid $x,y,z\ge0$ , $x+y+z\le1$ has bounds $0\le x\le1$ , $0\le y\le1-x$ , $0\le z\le1-x-y$ . Its volume is

\int_0^1\int_0^{1-x}\int_0^{1-x-y}1\,dz\,dy\,dx=\int_0^1\frac{(1-x)^2}{2}dx=\frac16.

For fixed $x$ , the remaining slice is a right triangle, whose area is $(1-x)^2/2$ . This geometric reading explains the intermediate integral.

Worked example: between two surfaces

Take $0\le x,y\le1$ and $x+y\le z\le x+y+2$ . Every vertical column has height two, so volume is $2$ . The integral of $z$ is

\int_0^1\int_0^1\frac{(x+y+2)^2-(x+y)^2}{2}dy dx=4.

The average height coordinate is $4/2=2$ . It lies between the smallest and largest heights in the solid, a useful sanity check.

Changing integration order may require splitting a solid into several regions. Start from the geometric inequalities and project again for the new outer variables. Do not merely permute the differential symbols. For continuous integrands over bounded regular solids, Fubini justifies the iterated evaluation; singular cases need convergence analysis.

Practice

Find the volume of $[0,2]\times[0,3]\times[0,4]$ .
Write bounds for $x,y,z\ge0$ and $x+y+z\le2$ with $z$ integrated first.
Integrate $x$ over the unit cube.

Show worked solutions

$2\cdot3\cdot4=24$ .
$0\le x\le2$ , $0\le y\le2-x$ , $0\le z\le2-x-y$ .
$\int_0^1x dx$ times the unit areas in the other coordinates gives $1/2$ .

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.