Iterated Integrals and Area in the Plane Explained

What Iterated Integrals Actually Are

An iterated integral is just what it sounds like: integration performed sequentially, one variable at a time. You evaluate one integral, then plug the result into another integral.

Most students encounter them when calculating double integrals over rectangular regions. The notation looks intimidating at first, but the concept is straightforward once you see it in action.

The Two Main Forms

When you want to find the area under a surface or calculate a double integral over a region R, you have two equivalent ways to write the computation:

Form 1: Integrate with respect to x first, then y

Form 2: Integrate with respect to y first, then x

Fubini's Theorem guarantees both orders give the same answer when the function is continuous over the region. If the function has discontinuities, you're in for a headache.

Iterated Integrals for Area in the Plane

For calculating area in the plane, iterated integrals strip down to something simpler. You can find the area of a region bounded by curves using:

• The double integral of 1 over the region: ∬_R dA
• Integrating the top function minus the bottom function
• Integrating the right function minus the left function

All three approaches work. The third option is where iterated integrals shine.

Example: Area Between Two Curves

Find the area bounded by y = x² and y = x + 2.

First, find where they intersect:

x² = x + 2 → x² - x - 2 = 0 → (x-2)(x+1) = 0 → x = 2 or x = -1

The area is:

A = ∫_-1² [(x + 2) - x²] dx

= [½x² + 2x - ⅓x³]_-1²

= (2 + 4 - 8/3) - (½ - 2 + ⅓)

= (6 - 8/3) - (-3/2 + ⅓)

= 10/3 - (-7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2

Double Integrals Over Rectangular Regions

For a rectangle [a,b] × [c,d], the iterated integral looks like:

∬_R f(x,y) dA = ∫_c^d ∫_a^b f(x,y) dx dy = ∫_a^b ∫_c^d f(x,y) dy dx

You can swap the order freely for rectangular regions. For non-rectangular regions, the limits change and you can't always swap without recalculating.

Example: Evaluating a Double Integral

Evaluate ∫₀² ∫₁³ (2x + y) dx dy

Step 1: Inner integral with respect to x:

∫₁³ (2x + y) dx = [x² + xy]₁³ = (9 + 3y) - (1 + y) = 8 + 2y

Step 2: Outer integral with respect to y:

∫₀² (8 + 2y) dy = [8y + y²]₀² = 16 + 4 = 20

Regions That Aren't Rectangles

This is where most people mess up. For a general region D in the xy-plane:

• Type I regions: bounded by curves x = g₁(y) and x = g₂(y) between y = c and y = d
• Type II regions: bounded by curves y = h₁(x) and y = h₂(x) between x = a and x = b

You must determine the region type before setting up your iterated integral. Draw it. I don't care if you think you can visualize it—draw the region.

Setting Up the Integral

For a Type I region, integrate x first:

∬_D f(x,y) dA = ∫_c^d ∫_g₁(y)^g₂(y) f(x,y) dx dy

For a Type II region, integrate y first:

∬_D f(x,y) dA = ∫_a^b ∫_h₁(x)^h₂(x) f(x,y) dy dx

When to Integrate in Which Order

Sometimes one order is significantly easier. Consider f(x,y) = xe^xy over a region.

If you integrate x first: ∫ xe^xy dx = requires substitution with y treated as constant.

If you integrate y first: ∫ xe^xy dy = x · (e^xy/x) = e^xy + C. Much simpler.

When given a choice, integrate the variable that makes the inside integral easier. This comes with practice.

Comparing Methods for Finding Area

Getting Started: Step-by-Step

Here's how to set up any iterated integral for area:

1. Sketch the region

Find intersection points. Plot them. Connect the dots. Know what shape you're dealing with.

2. Determine the region type

Can you describe x as a function of y? Type I. Can you describe y as a function of x? Type II.

3. Set the limits

For Type I: y goes from bottom to top (constants), x goes from left curve to right curve (functions of y).

For Type II: x goes from left to right (constants), y goes from bottom curve to top curve (functions of x).

4. Choose your integration order

Pick whichever makes the inner integral less painful. If both look bad, consider a coordinate transformation.

5. Evaluate

Work from the inside out. The inner integral gives you a function of the outer variable. Then integrate normally.

Common Mistakes

• Getting limits backwards—this flips the sign
• Using the wrong function for the boundary—check your sketch
• Forgetting to include all parts of a region if it's not simply connected
• Treating the variable incorrectly in the inner integral

The most common error is mixing up which variable is the function and which is the constant when setting limits. This comes from not drawing the region.

When You Need Triple Integrals

If you're working in three dimensions—finding volume, center of mass, or moments of inertia—you'll need triple integrals. The same iterated structure applies, just with an extra integration.

∭_E f(x,y,z) dV = ∫ ∫ ∫ f(x,y,z) dx dy dz

The order can be any permutation of dx dy dz. The limits get more complicated, but the principle stays the same.