05 Multiple Integrals

A structured guide to setting up and evaluating double and triple integrals, then applying them to area, volume, mass, average value, moments, and center of mass.

The role of multiple integrals

A multiple integral extends the definite integral to functions of two or three variables. The dimension of the region determines the type of integral:

A accumulates a quantity across a planar region.
A accumulates a quantity throughout a solid region.
Integrating the constant function $1$ gives area in two dimensions or volume in three dimensions.
Integrating a density gives mass.
Dividing total accumulation by area or volume gives an .

The central setup task is to describe the region correctly and choose bounds that include every point exactly once.

Takeaway: Identify the region and the quantity being accumulated before writing any bounds.

Double integrals over rectangular regions

For a rectangle $R=[a,b]\times[c,d]$ , the of $f(x,y)$ can be evaluated in either order. The two iterated forms are

\iint_R f(x,y)\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy.

This is an application of . The inner integration is performed first, while the other variable is treated as a constant.

For example,

\begin{aligned} \int_0^2\int_1^3 (x+2y)\,dy\,dx &=\int_0^2\left[xy+y^2\right]_1^3\,dx\\ &=\int_0^2(2x+8)\,dx\\ &=20. \end{aligned}

On a rectangle, constant bounds make either order valid. Choose the order that produces the easier inner integral.

Takeaway: Integrate the inner variable first and treat all remaining variables as constants during that step.

Describing general planar regions

For a nonrectangular region, the bounds must reflect the boundary curves. Begin with a sketch, then decide whether vertical or horizontal slices give the clearer description.

A has the form

D=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\},

so its integral is

\iint_D f(x,y)\,dA =\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx.

A has the form

D=\{(x,y):c\le y\le d,\ h_1(y)\le x\le h_2(y)\},

so its integral is

\iint_D f(x,y)\,dA =\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)\,dx\,dy.

If one description requires different bounds on different portions of the region, split the region into pieces and add the resulting integrals.

For the triangular region defined by $x\ge0$ , $y\ge0$ , and $x+y\le4$ , vertical slices give $0\le x\le4$ and $0\le y\le4-x$ . The volume under $z=4-x-y$ is

\begin{aligned} V &=\int_0^4\int_0^{4-x}(4-x-y)\,dy\,dx\\ &=\frac{1}{2}\int_0^4(4-x)^2\,dx\\ &=\frac{32}{3}. \end{aligned}

Takeaway: The outer bounds describe the full range of the outer variable; the inner bounds describe each slice between its boundary curves.

Applications of double integrals

The integrand determines what a measures.

Area:

\operatorname{Area}(D)=\iint_D1\,dA.

Volume under a surface: If $f(x,y)\ge0$ , then $V=\iint_D f(x,y)\,dA$ .
Volume between surfaces: If the upper surface is $f$ and the lower surface is $g$ , then $V=\iint_D\bigl(f(x,y)-g(x,y)\bigr)\,dA$ .
Mass of a lamina: With surface density $\rho(x,y)$ , $m=\iint_D\rho(x,y)\,dA$ .
:

f_{\mathrm{avg}}=\frac{1}{\operatorname{Area}(D)}\iint_Df(x,y)\,dA.

The denominator in an average-value formula is essential. The integral alone gives a total accumulation, not an average.

For the rectangle $D=[0,2]\times[1,4]$ ,

\operatorname{Area}(D)=\int_0^2\int_1^4 1\,dy\,dx=6.

If the density is constant, $\rho(x,y)=\rho_0$ , then the mass is $m=\rho_0\operatorname{Area}(D)$ .

Takeaway: Use $1$ for geometric measure, height for volume, density for mass, and the target function for an accumulation or average numerator.

Triple integrals and solid regions

A over a solid $E$ is written as $\iiint_E f(x,y,z)\,dV$ . For a box $B=[a,b]\times[c,d]\times[e,f]$ , one possible order is

\iiint_B f(x,y,z)\,dV =\int_e^f\int_c^d\int_a^b f(x,y,z)\,dx\,dy\,dz.

There are six possible orders. When the hypotheses of hold, all valid orders have the same value, although their bounds and computational difficulty can differ.

The volume of a solid is obtained by integrating $1$ :

\operatorname{Vol}(E)=\iiint_E1\,dV.

For the box $B=[0,2]\times[1,3]\times[-1,1]$ ,

\operatorname{Vol}(B)=\int_{-1}^{1}\int_1^3\int_0^2 1\,dx\,dy\,dz=8.

A solid can also be described by a planar base $D$ and lower and upper surfaces:

E=\{(x,y,z):(x,y)\in D,\ u(x,y)\le z\le v(x,y)\}.

Then

\iiint_E f(x,y,z)\,dV =\iint_D\int_{u(x,y)}^{v(x,y)}f(x,y,z)\,dz\,dA.

For the first-octant solid below $z=6-2x-3y$ , the projection is $0\le x\le3$ , $0\le y\le\frac{6-2x}{3}$ , and the volume is

V=\int_0^3\int_0^{(6-2x)/3}\int_0^{6-2x-3y}1\,dz\,dy\,dx=6.

Takeaway: For a solid between two surfaces, project onto a coordinate plane and integrate from the lower surface to the upper surface.

Mass, averages, and

Triple integrals extend the same applications into three dimensions.

Volume:

\operatorname{Vol}(E)=\iiint_E1\,dV.

Mass: For volume density $\rho(x,y,z)$ , $m=\iiint_E\rho(x,y,z)\,dV$ .
:

f_{\mathrm{avg}}=\frac{1}{\operatorname{Vol}(E)}\iiint_Ef(x,y,z)\,dV.

For a lamina with surface density $\rho(x,y)$ , the first moments are

M_x=\iint_D y\rho(x,y)\,dA, \qquad M_y=\iint_D x\rho(x,y)\,dA.

The total mass is $m=\iint_D\rho(x,y)\,dA$ , so the is

\bar{x}=\frac{M_y}{m}, \qquad \bar{y}=\frac{M_x}{m}.

For a solid, the corresponding moments are

M_{yz}=\iiint_E x\rho\,dV, \qquad M_{xz}=\iiint_E y\rho\,dV, \qquad M_{xy}=\iiint_E z\rho\,dV.

The coordinates of the are therefore

\bar{x}=\frac{M_{yz}}{m}, \qquad \bar{y}=\frac{M_{xz}}{m}, \qquad \bar{z}=\frac{M_{xy}}{m}.

Takeaway: Moments weight mass by distance from coordinate planes; dividing each moment by total mass gives the corresponding center-of-mass coordinate.

A reliable setup and checking strategy

Use the following sequence whenever you set up a multiple integral.

Identify the quantity. Use $1$ for area or volume, density for mass, and the given function for a total accumulation.
Sketch the region. Mark boundary curves, planes, intercepts, and the relevant projection.
Choose the projection. For a , select the coordinate plane that gives the simplest base region and vertical bounds.
Choose the order. Inner bounds may depend on variables integrated later, but outer bounds cannot depend on variables integrated earlier.
Write the bounds geometrically. Check that lower bounds do not exceed upper bounds and that every point is included exactly once.
Integrate one variable at a time. During each step, treat the remaining variables as constants.
Check the result. Confirm the sign, units, scale, and whether the answer matches the geometry.

For volume between surfaces, verify that the integrand is upper height minus lower height. For an average, verify that total accumulation has been divided by area or volume. For mass, verify that density is being integrated over the correct region.

Final takeaway: Most setup errors come from an incorrect region description. A careful sketch followed by bounds written in the chosen order makes the calculation reliable.