The Central Idea
In one-variable calculus, a definite integral
\[ \int_a^b f(x)\,dx \]accumulates contributions across an interval.
In two variables, a double integral
\[ \boxed{ \iint_R f(x,y)\,dA } \]accumulates contributions across a region \(R\) in the plane.
In three variables, a triple integral
\[ \boxed{ \iiint_E f(x,y,z)\,dV } \]accumulates contributions throughout a solid region \(E\).
From Riemann Sums to Double Integrals
Divide a planar region into many small pieces with areas
\[ \Delta A_1,\Delta A_2,\ldots. \]Choose a sample point
\[ (x_i^*,y_i^*) \]in each piece.
A two-dimensional Riemann sum is
\[ \sum_i f(x_i^*,y_i^*)\Delta A_i. \]In the limiting process, this becomes
\[ \boxed{ \iint_R f(x,y)\,dA }. \]The Area Element \(dA\)
The symbol
\[ dA \]represents an infinitesimal element of area.
In Cartesian coordinates it is commonly written as
\[ \boxed{ dA=dx\,dy } \]or
\[ \boxed{ dA=dy\,dx }. \]The order depends on how the region is described and how the iterated integral is organized.
Double Integral over a Rectangle
Suppose
\[ R=[a,b]\times[c,d]. \]Then
\[ \iint_R f(x,y)\,dA \]can usually be computed as an iterated integral:
\[ \boxed{ \int_a^b \int_c^d f(x,y)\,dy\,dx }. \]Reverse Order on a Rectangle
For the same rectangular region,
\[ R=[a,b]\times[c,d], \]we may also write
\[ \boxed{ \int_c^d \int_a^b f(x,y)\,dx\,dy }. \]Under standard continuity or integrability assumptions, both orders give the same value.
Fubini's Theorem
Fubini's theorem provides the theoretical foundation for evaluating many multiple integrals as repeated one-variable integrals.
For a continuous function on a rectangle,
\[ \boxed{ \iint_R f(x,y)\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy }. \]How Iterated Integration Works
In
\[ \int_a^b \int_c^d f(x,y)\,dy\,dx, \]the inner integral is taken first.
While integrating with respect to \(y\), treat \(x\) as a constant.
The result becomes a function of \(x\), which is then integrated with respect to \(x\).
Example: Double Integral over a Rectangle
Evaluate
\[ \int_0^2\int_1^3(x+2y)\,dy\,dx. \]Integrate first with respect to \(y\):
\[ \int_1^3(x+2y)\,dy = \left[xy+y^2\right]_1^3. \]Therefore,
\[ = (3x+9)-(x+1) = 2x+8. \]Complete the Rectangle Example
Now integrate with respect to \(x\):
\[ \int_0^2(2x+8)\,dx = \left[x^2+8x\right]_0^2. \]Thus,
\[ \boxed{ \int_0^2\int_1^3(x+2y)\,dy\,dx = 20 }. \]Constant Integrand and Area
If
\[ f(x,y)=1, \]then
\[ \boxed{ \iint_R 1\,dA = \operatorname{Area}(R) }. \]The double integral of \(1\) over a region measures its area.
Example: Area of a Rectangle
Let
\[ R=[0,3]\times[1,5]. \]Then
\[ \iint_R1\,dA = \int_0^3\int_1^5 1\,dy\,dx. \]The inner integral gives \(4\), so
\[ \int_0^3 4\,dx=12. \]Therefore,
\[ \boxed{ \operatorname{Area}(R)=12 }. \]Volume under a Surface
If
\[ f(x,y)\ge0 \]on \(R\), then
\[ \boxed{ V = \iint_R f(x,y)\,dA } \]gives the volume under the surface
\[ z=f(x,y) \]and above the region \(R\).
Signed Accumulation
If \(f\) takes both positive and negative values, then
\[ \iint_R f(x,y)\,dA \]represents signed accumulation.
Contributions above the reference plane count positively, while contributions below it count negatively.
General Planar Regions
Many regions are not rectangles.
A common type of region is bounded vertically:
\[ R = \{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\}. \]Then
\[ \boxed{ \iint_R f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx }. \]Horizontally Described Regions
A region may instead be written as
\[ R = \{(x,y):c\le y\le d,\ h_1(y)\le x\le h_2(y)\}. \]Then
\[ \boxed{ \iint_R f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy }. \]Reading Variable Limits
In
\[ \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx, \]the outer variable \(x\) ranges over the full horizontal span
\[ a\le x\le b. \]For each fixed \(x\), the inner variable \(y\) moves from the lower boundary \(g_1(x)\) to the upper boundary \(g_2(x)\).
Example: A Triangular Region
Let \(R\) be the triangle bounded by
\[ x=0, \qquad y=0, \qquad x+y=1. \]Solving for \(y\),
\[ y=1-x. \]Thus,
\[ 0\le x\le1, \qquad 0\le y\le1-x. \]Area of the Triangular Region
Use
\[ \operatorname{Area}(R) = \int_0^1\int_0^{1-x}1\,dy\,dx. \]The inner integral is
\[ 1-x. \]Therefore,
\[ \int_0^1(1-x)\,dx = \left[x-\frac{x^2}{2}\right]_0^1 = \frac12. \]Hence,
\[ \boxed{ \operatorname{Area}(R)=\frac12 }. \]Changing the Order of Integration
The same region can often be described in the opposite order.
For the triangle
\[ x\ge0,\qquad y\ge0,\qquad x+y\le1, \]we can also write
\[ 0\le y\le1, \qquad 0\le x\le1-y. \]Therefore,
\[ \boxed{ \int_0^1\int_0^{1-x}f(x,y)\,dy\,dx = \int_0^1\int_0^{1-y}f(x,y)\,dx\,dy }. \]Why Change the Order?
Reversing the order of integration may:
- simplify the antiderivative;
- replace difficult limits with simpler ones;
- make a previously impossible elementary integral manageable;
- fit the geometry of the region more naturally.
The Region Comes First
When changing the order, do not merely reverse the symbols \(dx\) and \(dy\).
First identify the geometric region.
Then describe that same region using the new order.
Example: Integrating over a Triangle
Evaluate
\[ \int_0^1\int_0^{1-x}(x+y)\,dy\,dx. \]First integrate with respect to \(y\):
\[ \int_0^{1-x}(x+y)\,dy = x(1-x)+\frac{(1-x)^2}{2}. \]Simplifying,
\[ = \frac12-\frac{x^2}{2}. \]Complete the Triangle Integral
Now integrate:
\[ \int_0^1 \left( \frac12-\frac{x^2}{2} \right)\,dx. \]Therefore,
\[ = \left[ \frac{x}{2} - \frac{x^3}{6} \right]_0^1 = \frac13. \]Hence,
\[ \boxed{ \int_0^1\int_0^{1-x}(x+y)\,dy\,dx = \frac13 }. \]Average Value over a Region
The average value of \(f\) over a region \(R\) of finite positive area is
\[ \boxed{ f_{\text{avg}} = \frac{1}{\operatorname{Area}(R)} \iint_R f(x,y)\,dA }. \]Example: Average Height
Suppose
\[ h(x,y)=x+y \]on the unit square
\[ 0\le x\le1, \qquad 0\le y\le1. \]The area is \(1\), so
\[ h_{\text{avg}} = \int_0^1\int_0^1(x+y)\,dy\,dx. \]The integral equals \(1\), giving
\[ \boxed{ h_{\text{avg}}=1 }. \]Mass of a Lamina
Suppose a thin planar object occupies a region \(R\) and has density
\[ \rho(x,y). \]Its mass is
\[ \boxed{ M = \iint_R\rho(x,y)\,dA }. \]Constant Density
If the density is constant,
\[ \rho(x,y)=\rho_0, \]then
\[ M = \rho_0 \iint_R1\,dA. \]Therefore,
\[ \boxed{ M = \rho_0\operatorname{Area}(R) }. \]Moments and Center of Mass
For a lamina with density \(\rho(x,y)\), the first moments are
\[ M_x = \iint_R y\rho(x,y)\,dA \]and
\[ M_y = \iint_R x\rho(x,y)\,dA. \]If total mass is \(M\), then the center of mass is
\[ \boxed{ \bar{x}=\frac{M_y}{M}, \qquad \bar{y}=\frac{M_x}{M} }. \]Probability Density in Two Variables
If
\[ p(x,y) \]is a joint probability density, then probabilities are computed by integrating over regions:
\[ \boxed{ P((X,Y)\in R) = \iint_R p(x,y)\,dA }. \]The total density must satisfy
\[ \iint_{\mathbb{R}^2}p(x,y)\,dA=1. \]Expectation from a Joint Density
For suitable random variables with joint density \(p(x,y)\),
\[ E[X] = \iint_{\mathbb{R}^2} x\,p(x,y)\,dA, \]and
\[ E[Y] = \iint_{\mathbb{R}^2} y\,p(x,y)\,dA. \]Multiple integration therefore connects directly to probability and statistics.
Triple Integrals
A triple integral accumulates a quantity throughout a three-dimensional region \(E\):
\[ \boxed{ \iiint_E f(x,y,z)\,dV }. \]Here
\[ dV \]represents an infinitesimal volume element.
Cartesian Volume Element
In Cartesian coordinates,
\[ \boxed{ dV=dx\,dy\,dz } \]in one possible order.
Other orders such as
\[ dz\,dy\,dx \]are equally valid when the bounds are adjusted to describe the same solid.
Triple Integral over a Box
Suppose
\[ E = [a,b]\times[c,d]\times[e,g]. \]Then a triple integral may be written
\[ \boxed{ \int_a^b \int_c^d \int_e^g f(x,y,z)\,dz\,dy\,dx }. \]Volume of a Solid
Just as a double integral of \(1\) gives area,
\[ \boxed{ \iiint_E1\,dV = \operatorname{Vol}(E) }. \]Example: Volume of a Box
Let
\[ E=[0,2]\times[1,4]\times[-1,1]. \]Then
\[ \iiint_E1\,dV = \int_0^2\int_1^4\int_{-1}^1 1\,dz\,dy\,dx. \]The side lengths are \(2\), \(3\), and \(2\), so
\[ \boxed{ \operatorname{Vol}(E)=12 }. \]Mass of a Three-Dimensional Body
If a solid \(E\) has density
\[ \rho(x,y,z), \]then its mass is
\[ \boxed{ M = \iiint_E \rho(x,y,z)\,dV }. \]Average Value over a Solid
If \(E\) has finite positive volume, then
\[ \boxed{ f_{\text{avg}} = \frac{1}{\operatorname{Vol}(E)} \iiint_E f(x,y,z)\,dV }. \]Describing a Solid with Variable Bounds
A general solid may be described by nested inequalities such as
\[ a\le x\le b, \] \[ g_1(x)\le y\le g_2(x), \] \[ h_1(x,y)\le z\le h_2(x,y). \]Then
\[ \boxed{ \iiint_E f\,dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f(x,y,z)\,dz\,dy\,dx }. \]Projection of a Solid
When integrating in the order
\[ dz\,dy\,dx, \]the outer two variables describe the projection of the solid onto the \(xy\)-plane.
For each \((x,y)\) in that projection, the inner \(z\)-bounds describe the bottom and top surfaces.
Example: Solid under a Plane
Consider the solid in the first octant beneath
\[ z=1-x-y. \]Since \(z\ge0\),
\[ x+y\le1. \]The projection onto the \(xy\)-plane is the triangle
\[ 0\le x\le1, \qquad 0\le y\le1-x. \]For each point in that triangle,
\[ 0\le z\le1-x-y. \]Triple-Integral Setup
The volume is therefore
\[ \boxed{ V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1\,dz\,dy\,dx }. \]Complete the Volume Calculation
Integrate first with respect to \(z\):
\[ \int_0^{1-x-y}1\,dz = 1-x-y. \]Then
\[ V = \int_0^1 \int_0^{1-x} (1-x-y)\,dy\,dx. \]Evaluating gives
\[ \boxed{ V=\frac16 }. \]Symmetry Can Simplify Integrals
If a region is symmetric and the integrand has an odd symmetry, the integral may vanish.
For example, if \(R\) is symmetric about the \(y\)-axis and
\[ f(-x,y)=-f(x,y), \]then, under appropriate integrability assumptions,
\[ \boxed{ \iint_R f(x,y)\,dA=0 }. \]Separating Products on Rectangles
If
\[ f(x,y)=g(x)h(y) \]and
\[ R=[a,b]\times[c,d], \]then
\[ \boxed{ \iint_R g(x)h(y)\,dA = \left( \int_a^b g(x)\,dx \right) \left( \int_c^d h(y)\,dy \right) }. \]The rectangular geometry is essential to this simple product factorization.
Example: Product Separation
Evaluate
\[ \int_0^1\int_0^2 xe^y\,dy\,dx. \]Because the region is rectangular and the integrand separates,
\[ = \left(\int_0^1x\,dx\right) \left(\int_0^2e^y\,dy\right). \]Therefore,
\[ \boxed{ \frac12(e^2-1) }. \]Units of a Double Integral
If \(f\) has units \(U\), while \(x\) and \(y\) each have units of length, then
\[ f\,dA \]has units
\[ U\cdot L^2. \]The resulting integral has the same units.
Units of a Triple Integral
If \(f\) has units \(U\), then
\[ f\,dV \]has units
\[ U\cdot L^3 \]when the three spatial coordinates are measured in units of length.
Density Units
A surface density
\[ \rho(x,y) \]has units such as
\[ \frac{\text{mass}}{\text{area}}. \]Multiplying by \(dA\) leaves units of mass.
A volume density
\[ \rho(x,y,z) \]has units such as
\[ \frac{\text{mass}}{\text{volume}}. \]Multiple Integrals as Weighted Sums
The common structure is
\[ \boxed{ \text{accumulated quantity} = \int (\text{local density or contribution}) \times (\text{small region element}) }. \]This viewpoint unifies area, volume, mass, probability, and average-value calculations.
Numerical Multiple Integration
Not every multiple integral has a convenient symbolic antiderivative.
Numerical methods approximate the integral by finite weighted sums, quadrature rules, sampling methods, or computational algorithms.
The underlying principle remains the same: partition the region and add local contributions.
Why Geometry Matters So Much
For multiple integrals, the integrand is only part of the problem.
The region of integration determines:
- the bounds;
- the useful order of integration;
- possible symmetry;
- whether a coordinate change may simplify the calculation.
Drawing or mentally organizing the region is often the most important first step.
When Cartesian Coordinates Are Awkward
A circular region such as
\[ x^2+y^2\le9 \]can be described in Cartesian coordinates, but the limits involve square roots.
A coordinate system adapted to circles can make the same region much simpler.
This motivates coordinate changes and Jacobians.
Common Error: Integrating in the Wrong Order
In
\[ \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx, \]the \(y\)-integration must be performed first.
Read nested integrals from the inside outward.
Common Error: Treating the Other Variable as Variable Too Soon
During
\[ \int f(x,y)\,dy, \]the variable \(x\) is treated as a constant.
This is the same operational principle used in partial differentiation.
Common Error: Using Bounds That Depend on the Inner Variable
In an iterated integral, the limits of the inner variable may depend on variables that are integrated later, but not on the inner variable itself.
For example,
\[ \int_0^1\int_0^{1-x}f(x,y)\,dy\,dx \]is structurally valid because the \(y\)-bounds depend on \(x\), the outer variable.
Common Error: Reversing Order without Redrawing the Region
The expression
\[ \int_0^1\int_0^{1-x}f(x,y)\,dy\,dx \]cannot be reversed simply by swapping \(dx\) and \(dy\).
The bounds must be rewritten to describe the same geometric region.
Common Error: Confusing Area and Volume
The integral
\[ \iint_R1\,dA \]gives area.
The integral
\[ \iint_Rf(x,y)\,dA \]can give volume when \(f\ge0\).
The integral
\[ \iiint_E1\,dV \]gives three-dimensional volume.
Common Error: Forgetting Density
If density varies, mass is not generally
\[ \operatorname{Area}(R) \]or
\[ \operatorname{Vol}(E). \]The density must be included:
\[ M=\iint_R\rho\,dA \]or
\[ M=\iiint_E\rho\,dV. \]A Reliable Double-Integral Procedure
- Identify or sketch the region \(R\).
- Choose an order of integration.
- Write the outer bounds for the full span of the outer variable.
- Write the inner bounds across the region for each fixed outer-variable value.
- Integrate from the inside outward.
- Check units and geometric reasonableness.
A Reliable Triple-Integral Procedure
- Understand the solid \(E\).
- Choose an integration order.
- Identify the projection associated with the outer variables.
- Write the innermost bounds between the relevant bounding surfaces.
- Integrate one variable at a time.
- Check the result against the geometry and units.
Example: Full Double-Integral Workflow
Evaluate
\[ \iint_R xy\,dA \]over the region
\[ 0\le x\le2, \qquad 0\le y\le x. \]The iterated integral is
\[ \int_0^2\int_0^xxy\,dy\,dx. \]Complete the Full Double-Integral Workflow
Integrate with respect to \(y\):
\[ \int_0^xxy\,dy = x\left[\frac{y^2}{2}\right]_0^x = \frac{x^3}{2}. \]Then
\[ \int_0^2\frac{x^3}{2}\,dx = \frac12 \left[ \frac{x^4}{4} \right]_0^2. \]Therefore,
\[ \boxed{ \iint_Rxy\,dA=2 }. \]Practice 1: Rectangle
Evaluate
\[ \int_0^1\int_0^2(x+y)\,dy\,dx. \]Practice 2: Area by Double Integration
Use a double integral to find the area of
\[ 0\le x\le2, \qquad 0\le y\le3. \]Practice 3: Triangular Region
Evaluate
\[ \int_0^1\int_0^{1-x}1\,dy\,dx. \]Practice 4: Variable Integrand
Evaluate
\[ \int_0^1\int_0^x y\,dy\,dx. \]Practice 5: Reverse the Order
Rewrite
\[ \int_0^1\int_0^{1-x} f(x,y)\,dy\,dx \]in the order \(dx\,dy\).
Practice 6: Average Value
Find the average value of
\[ f(x,y)=x+y \]on
\[ 0\le x\le2, \qquad 0\le y\le2. \]Practice 7: Mass of a Lamina
A rectangular lamina occupies
\[ 0\le x\le1, \qquad 0\le y\le2 \]and has density
\[ \rho(x,y)=x+y. \]Find its mass.
Practice 8: Triple Integral over a Box
Evaluate
\[ \int_0^1\int_0^2\int_0^3 1\,dz\,dy\,dx. \]Practice 9: Triple Integral with Integrand
Evaluate
\[ \int_0^1\int_0^1\int_0^1 (x+y+z)\,dz\,dy\,dx. \]Practice 10: Identify the Error
A student tries to reverse
\[ \int_0^1\int_0^{1-x} f(x,y)\,dy\,dx \]by writing
\[ \int_0^1\int_0^{1-x} f(x,y)\,dx\,dy. \]Explain the error.
Practice Answers
Practice 1
First integrate with respect to \(y\):
\[ \int_0^2(x+y)\,dy = 2x+2. \]Then
\[ \int_0^1(2x+2)\,dx = 3. \]Therefore,
\[ \boxed{3}. \]Practice 2
\[ \operatorname{Area} = \int_0^2\int_0^3 1\,dy\,dx. \]Therefore,
\[ \boxed{6}. \]Practice 3
\[ \int_0^1\int_0^{1-x}1\,dy\,dx = \int_0^1(1-x)\,dx = \boxed{\frac12}. \]Practice 4
\[ \int_0^x y\,dy = \frac{x^2}{2}. \]Then
\[ \int_0^1\frac{x^2}{2}\,dx = \boxed{\frac16}. \]Practice 5
The region is
\[ x\ge0, \qquad y\ge0, \qquad x+y\le1. \]In the order \(dx\,dy\),
\[ \boxed{ \int_0^1 \int_0^{1-y} f(x,y)\,dx\,dy }. \]Practice 6
The region has area
\[ 4. \]Also,
\[ \int_0^2\int_0^2(x+y)\,dy\,dx = 8. \]Therefore,
\[ \boxed{ f_{\text{avg}}=\frac84=2 }. \]Practice 7
The mass is
\[ M = \int_0^1\int_0^2(x+y)\,dy\,dx. \]The inner integral is
\[ 2x+2. \]Therefore,
\[ M = \int_0^1(2x+2)\,dx = \boxed{3}. \]Practice 8
The integral gives the volume of the box:
\[ \boxed{ 1\cdot2\cdot3=6 }. \]Practice 9
By symmetry or direct integration, the average contribution from each variable over \([0,1]\) is \(1/2\).
Therefore,
\[ \boxed{ \int_0^1\int_0^1\int_0^1 (x+y+z)\,dz\,dy\,dx = \frac32 }. \]Practice 10
The student changed the differential order without rewriting the bounds. The bounds
\[ 0\le y\le1-x \]describe \(y\) in terms of \(x\).
For the order \(dx\,dy\), the same triangular region must be written
\[ 0\le y\le1, \qquad 0\le x\le1-y. \]Thus the correct reversed integral is
\[ \boxed{ \int_0^1\int_0^{1-y} f(x,y)\,dx\,dy }. \]What to Remember
Multiple integration extends accumulation from intervals to regions and solids.
- A double integral accumulates over a planar region.
- A triple integral accumulates over a three-dimensional solid.
- Iterated integrals are evaluated from the inside outward.
- The inner bounds describe a slice of the region for each fixed value of the outer variable or variables.
- Integrating \(1\) gives area in two dimensions and volume in three.
- Integrating density gives mass or probability.
- Dividing accumulated value by area or volume gives an average value.
- Changing the order requires redescribing the same geometric region.
The unifying structure is
\[ \boxed{ \text{total accumulation} = \int \text{local contribution} \times \text{small geometric element} }. \]Where We Go Next
Cartesian coordinates are effective for rectangular and many graph-bounded regions, but they can become cumbersome when the geometry has circles, ellipses, radial symmetry, or other natural coordinate structures.
The next lesson, Coordinate Changes and Jacobians, develops systematic methods for replacing one coordinate system with another and explains how area and volume elements must scale under those transformations.