The Main Idea 

Single integrals find areas under curves. Double integrals extend this idea to find volumes under surfaces. Instead of slicing a region into thin strips, you’re dividing a rectangular base into small boxes and stacking them up to the surface.

For a function [latex]z = f(x, y)[/latex] over a rectangular region [latex]R = [a, b] \times [c, d][/latex], the double integral is:

[latex]\large{\underset{R}{\displaystyle\iint}f(x,y)dA=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^{m}\sum_{j=1}^{n}f(x_i^*,y_j^*)\Delta A}[/latex].

This represents the volume of the solid sitting above region [latex]R[/latex] and below the surface [latex]z = f(x, y)[/latex].

As you make the grid finer (more and smaller rectangles), the approximation improves. The exact volume is the limit as [latex]m[/latex] and [latex]n[/latex] approach infinity—that limit is the double integral.

Key Notation

Since [latex]dA = dx,dy = dy,dx[/latex], you can write the double integral as:

[latex]\underset{R}{\displaystyle\iint}f(x,y)dx,dy[/latex] or [latex]\underset{R}{\displaystyle\iint}f(x,y)dy,dx[/latex]

When It Works

The double integral exists if [latex]f[/latex] is continuous over [latex]R[/latex] (or at least not too discontinuous). If [latex]f(x, y) \geq 0[/latex], the result is volume. If [latex]f[/latex] is sometimes negative, think of it as signed volume—parts below the [latex]xy[/latex]-plane count as negative.

The Main Idea 

Double integrals behave much like single integrals—they follow familiar algebraic rules that make calculations easier. Understanding these properties helps you break complex integrals into simpler pieces.

Linearity (Properties 1 and 2)

Double integrals respect addition and scalar multiplication:

[latex]\underset{R}{\displaystyle\iint}[f(x,y)+g(x,y)]dA=\underset{R}{\displaystyle\iint}f(x,y)dA+\underset{R}{\displaystyle\iint}g(x,y)dA[/latex]

[latex]\underset{R}{\displaystyle\iint}cf(x,y)dA = c\underset{R}{\displaystyle\iint}f(x,y)dA[/latex]

You can split integrals of sums into separate integrals, and pull constants outside the integral. This works exactly like it did in single-variable calculus.

Additivity (Property 3)

If you split region [latex]R[/latex] into non-overlapping subregions [latex]S[/latex] and [latex]T[/latex], the integral over [latex]R[/latex] equals the sum of integrals over the pieces:

[latex]\underset{R}{\displaystyle\iint}f(x,y)dA = \underset{S}{\displaystyle\iint}f(x,y)dA + \underset{T}{\displaystyle\iint}f(x,y)dA[/latex]

This is useful when your region has a natural division or when part is easier to integrate than the whole.

Monotonicity (Property 4)

If [latex]f(x,y) \geq g(x,y)[/latex] everywhere on [latex]R[/latex], then their integrals follow the same inequality. Bigger functions produce bigger integrals.

Bounding the Integral (Property 5)

If you know [latex]m \leq f(x,y) \leq M[/latex] on region [latex]R[/latex], then:

[latex]m \times A(R) \leq \underset{R}{\displaystyle\iint}f(x,y)dA \leq M \times A(R)[/latex]

This gives quick estimates: multiply the region’s area by the minimum and maximum function values.

Separable Functions (Property 6)

If [latex]f(x,y) = g(x)h(y)[/latex] (a product of separate [latex]x[/latex] and [latex]y[/latex] functions), the double integral factors:

[latex]\underset{R}{\displaystyle\iint}g(x)h(y)dA = \left(\displaystyle\int_a^bg(x)dx\right)\left(\displaystyle\int_c^dh(y)dy\right)[/latex]

This turns a double integral into two single integrals—often much easier to compute.

The Main Idea 

Computing double integrals directly from the limit definition would be painfully tedious. Fortunately, Fubini’s theorem gives us a practical method: break a double integral into two single integrals, integrating one variable at a time.

The Key Idea: Iterated Integrals

An iterated integral lets you evaluate a double integral by integrating twice—first with respect to one variable (treating the other as constant), then with respect to the second variable:

[latex]\displaystyle\int_a^b\int_c^df(x,y)dy,dx = \displaystyle\int_a^b\left[\displaystyle\int_c^df(x,y)dy\right]dx[/latex]

The inner integral [latex]\int_c^d f(x,y)dy[/latex] treats [latex]x[/latex] as a constant and integrates with respect to [latex]y[/latex]. The result is a function of [latex]x[/latex] alone, which you then integrate.

Fubini’s Theorem

For a continuous function [latex]f(x,y)[/latex] over rectangle [latex]R = [a,b] \times [c,d][/latex]:

[latex]\underset{R}{\displaystyle\iint}f(x,y)dA = \displaystyle\int_a^b\displaystyle\int_c^df(x,y)dy,dx = \displaystyle\int_c^d\displaystyle\int_a^bf(x,y)dx,dy[/latex]

Both orders of integration give the same answer. Choose whichever order makes the computation easier.

Problem-Solving Strategy

1. Identify your region: Note the bounds [latex]a \leq x \leq b[/latex] and [latex]c \leq y \leq d[/latex]
2. Choose your order: Decide whether to integrate with respect to [latex]x[/latex] or [latex]y[/latex] first
3. Integrate the inner integral: Treat one variable as constant and integrate with respect to the other
4. Integrate the outer integral: Integrate the resulting expression with respect to the remaining variable

When Order Matters

Sometimes one order is much easier than the other. If the inner integral looks complicated (requiring integration by parts or substitution that depends on both variables), try switching the order. A function like [latex]x\sin(xy)[/latex] integrates more easily with respect to [latex]y[/latex] first than [latex]x[/latex] first.

Critical Reminder

The limits of integration come directly from the region boundaries. For [latex]R = [a,b] \times [c,d][/latex], [latex]x[/latex] ranges from [latex]a[/latex] to [latex]b[/latex] and [latex]y[/latex] ranges from [latex]c[/latex] to [latex]d[/latex], regardless of integration order.

The Main Idea 

Most real-world regions aren’t rectangles. To integrate over irregular shapes, we classify them as Type I or Type II regions based on how their boundaries are defined. This determines which order of integration works naturally.

For any bounded region [latex]D[/latex], we can enclose it in a rectangle [latex]R[/latex]. We extend our function [latex]f(x,y)[/latex] to be zero outside [latex]D[/latex], allowing us to use rectangular integration techniques while only “counting” the area inside [latex]D[/latex].

Type I Regions: Vertical Slicing

A Type I region sits between two vertical lines [latex]x = a[/latex] and [latex]x = b[/latex], with top boundary [latex]y = g_2(x)[/latex] and bottom boundary [latex]y = g_1(x)[/latex]:

[latex]D = {(x,y) \mid a \leq x \leq b, g_1(x) \leq y \leq g_2(x)}[/latex]

Think vertically: for each fixed [latex]x[/latex]-value, [latex]y[/latex] ranges between two functions of [latex]x[/latex].

Type II Regions: Horizontal Slicing

A Type II region sits between two horizontal lines [latex]y = c[/latex] and [latex]y = d[/latex], with left boundary [latex]x = h_1(y)[/latex] and right boundary [latex]x = h_2(y)[/latex]:

[latex]D = {(x,y) \mid c \leq y \leq d, h_1(y) \leq x \leq h_2(y)}[/latex]

Think horizontally: for each fixed [latex]y[/latex]-value, [latex]x[/latex] ranges between two functions of [latex]y[/latex].

How to Identify Region Type

• Type I: Draw vertical lines through the region. If each vertical line enters and exits the region just once (crossing top and bottom boundaries), it’s Type I.
• Type II: Draw horizontal lines through the region. If each horizontal line enters and exits just once (crossing left and right boundaries), it’s Type II.

Many regions can be described either way. The region between [latex]y = \sqrt{x}[/latex] and [latex]y = x^3[/latex] for [latex]0 \leq x \leq 1[/latex] is Type I with [latex]x^3 \leq y \leq \sqrt{x}[/latex]. Solve for [latex]x[/latex] to get Type II: [latex]y^2 \leq x \leq \sqrt[3]{y}[/latex] for [latex]0 \leq y \leq 1[/latex]. Choose whichever makes the integral easier to compute.

The Main Idea 

Once you’ve identified your region as Type I, Type II, or a combination, you can set up the double integral using Fubini’s theorem. The key is translating the region description into correct limits of integration.

Fubini’s Theorem for General Regions

For a Type I region where [latex]a \leq x \leq b[/latex] and [latex]g_1(x) \leq y \leq g_2(x)[/latex]:

[latex]\underset{D}{\displaystyle\iint}f(x,y)dA = \displaystyle\int_a^b\left[\displaystyle\int_{g_1(x)}^{g_2(x)}f(x,y)dy\right]dx[/latex]

For a Type II region where [latex]c \leq y \leq d[/latex] and [latex]h_1(y) \leq x \leq h_2(y)[/latex]:

[latex]\underset{D}{\displaystyle\iint}f(x,y)dA = \displaystyle\int_c^d\left[\displaystyle\int_{h_1(y)}^{h_2(y)}f(x,y)dx\right]dy[/latex]

How the Limits Work

The outer integral limits are constants (the fixed interval). The inner integral limits are functions that depend on the outer variable. For Type I, [latex]y[/latex] varies between two curves that depend on [latex]x[/latex]. For Type II, [latex]x[/latex] varies between two curves that depend on [latex]y[/latex].

Problem-Solving Strategy

1. Sketch the region and identify all boundary curves
2. Decide on region type: Choose Type I or Type II based on which gives simpler limits
3. Write the bounds: Outer limits are constants, inner limits are functions
4. Integrate: Work from the inside out, treating the outer variable as constant in the inner integral

Sometimes one order makes integration much harder. If the inner integral looks difficult or impossible with your first choice, try describing the region the other way and switching the order of integration.

Complex Regions: Break Them Apart

If your region can’t be described as a single Type I or Type II region, split it into pieces:

[latex]\underset{D}{\displaystyle\iint}f(x,y)dA = \underset{D_1}{\displaystyle\iint}f(x,y)dA + \underset{D_2}{\displaystyle\iint}f(x,y)dA[/latex]

Evaluate each piece separately using whichever type fits that piece best, then add the results.

The Main Idea 

Sometimes the order of integration dramatically affects how difficult the problem is to solve. Learning to reverse the order—switching from [latex]dy,dx[/latex] to [latex]dx,dy[/latex] or vice versa—is a valuable skill that can turn an impossible integral into a manageable one.

Why Change the Order?

An integral might be difficult or impossible to evaluate in one order but straightforward in the other. If you encounter an inner integral you can’t solve, or one that requires complex techniques, switching the order of integration often provides a simpler path.

How to Reverse the Order

1. Identify the current region type: Read the limits carefully. Is it Type I ([latex]dy,dx[/latex]) or Type II ([latex]dx,dy[/latex])?
2. Sketch the region: Draw the boundaries to visualize what you’re working with
3. Rewrite as the opposite type: Convert Type I to Type II or vice versa by expressing the same region with different bounds
4. Solve any equations needed: If the region is bounded by [latex]y = 2 - x^2[/latex], solve for [latex]x[/latex] to get [latex]x = \sqrt{2-y}[/latex] when converting
5. Set up the new integral: Write the limits in the new order

Reading the Original Limits

From [latex]\displaystyle\int_{x=a}^{x=b}\displaystyle\int_{y=g_1(x)}^{y=g_2(x)}f(x,y)dy,dx[/latex]:

• The region spans [latex]a \leq x \leq b[/latex] (constants tell you the outer variable’s range)
• For each [latex]x[/latex], [latex]y[/latex] goes from [latex]g_1(x)[/latex] to [latex]g_2(x)[/latex] (functions tell you the inner variable’s range)

Setting Up the New Limits

After sketching, determine:

• What constant values bound the new outer variable?
• What functions (of the new outer variable) bound the new inner variable?

Pro Tip

Always sketch the region when changing order. Visual confirmation prevents mistakes with limits and helps you see which boundaries to use.