Double Integral Examples and Problems- Practice Guide

Double Integrals: What They Actually Are

A double integral extends single-variable integration to functions of two variables. You integrate over a 2D region instead of an interval. That's the whole idea.

Mathematically, ∫∫_R f(x,y) dA gives you the volume under the surface z = f(x,y) over region R (when f ≥ 0), or a signed volume otherwise.

Iterated Integrals: The Basic Setup

You evaluate double integrals by performing two successive single integrations. The inner integral runs first, then the outer.

The Standard Form

∫_a^b ∫_c^d f(x,y) dy dx

This means you integrate with respect to y first (holding x constant), then integrate the result with respect to x.

Example 1: Rectangular Region

Evaluate ∫₀¹ ∫₀² (x + 2y) dy dx

Step 1: Inner integral with respect to y:

∫₀² (x + 2y) dy = [xy + y²]₀² = 2x + 4

Step 2: Outer integral with respect to x:

∫₀¹ (2x + 4) dx = [x² + 4x]₀¹ = 1 + 4 = 5

The answer is 5.

Defining the Region: Type I and Type II

Most double integral problems don't give you rectangular regions. You get weird shapes bounded by curves. There are two ways to describe them.

Type I Regions

Regions bounded between two functions of x:

a ≤ x ≤ b

g₁(x) ≤ y ≤ g₂(x)

Integral: ∫_a^b ∫_g₁(x)^g₂(x) f(x,y) dy dx

Type II Regions

Regions bounded between two functions of y:

c ≤ y ≤ d

h₁(y) ≤ x ≤ h₂(y)

Integral: ∫_c^d ∫_h₁(y)^h₂(y) f(x,y) dx dy

How to Set Up Any Double Integral

Follow this process every time:

1. Draw the region — Sketch the bounding curves
2. Find intersection points — Solve simultaneously to get limits
3. Choose integration order — Pick the simpler one
4. Write the limits — Based on your choice of Type I or II
5. Evaluate — Inner first, then outer

Example 2: Setting Up from a Region

Set up ∫∫ f(x,y) dA over the region bounded by y = x and y = x².

Step 1: Find intersections. x = x² gives x(x-1) = 0, so x = 0 and x = 1.

Step 2: Determine y-limits. The region is between y = x² (lower) and y = x (upper) for x from 0 to 1.

Type I setup:

∫₀¹ ∫_x²^x f(x,y) dy dx

Type II setup:

For each y, x goes from x = y to x = 1 (since y ≤ x ≤ √y for y in [0,1]).

∫₀¹ ∫_y^√y f(x,y) dx dy

Changing the Order of Integration

Sometimes the given order makes the integral impossible or messy. Switching can save you.

Example 3: Switching Orders

Evaluate ∫₀¹ ∫_y¹ e^x² dx dy

The inner integral ∫ e^x² dx has no elementary antiderivative. You cannot evaluate it directly.

Switch the order:

The region: x goes from y to 1, y goes from 0 to 1.

This means y ≤ x ≤ 1 and 0 ≤ y ≤ x.

Rewriting: 0 ≤ y ≤ x and 0 ≤ x ≤ 1

New integral: ∫₀¹ ∫₀^x e^x² dy dx

Evaluate:

∫₀¹ e^x² [y]₀^x dx = ∫₀¹ x e^x² dx

= (1/2) [e^x²]₀¹ = (1/2)(e - 1)

This works. The original didn't.

Applications of Double Integrals

Finding Area

Area of region R = ∫∫_R 1 dA

Just integrate 1 over the region.

Finding Volume

Volume under z = f(x,y) over R = ∫∫_R f(x,y) dA

For volume between two surfaces f(x,y) and g(x,y):

V = ∫∫_R [f(x,y) - g(x,y)] dA

Average Value

Average value of f over R = (1/Area of R) ∫∫_R f(x,y) dA

Polar Coordinates in Double Integrals

When the region or integrand involves x² + y², switch to polar. It makes everything simpler.

Conversion:

• x = r cos θ
• y = r sin θ
• x² + y² = r²
• dA = dx dy = r dr dθ

Example 4: Circular Region

Find the volume under z = 4 - x² - y² above the xy-plane.

The region: x² + y² ≤ 4, or r ≤ 2 in polar.

Volume = ∫∫ (4 - r²) dA = ∫₀^2π ∫₀² (4 - r²) r dr dθ

Evaluate:

∫₀² (4r - r³) dr = [2r² - r⁴/4]₀² = 8 - 4 = 4

∫₀^2π 4 dθ = 4(2π) = 8π

Volume = 8π

Common Mistakes to Avoid

Practice Problems

Problem 1: Evaluate ∫₁² ∫₀³ (2x - y) dy dx

Answer: 15

Problem 2: Set up and evaluate ∫∫ xy dA over the triangle with vertices (0,0), (1,0), (0,2).

Answer: 1/6

Problem 3: Evaluate ∫₀¹ ∫_x¹ sin(y²) dy dx by switching the order.

Answer: (1 - cos(1))/2

Problem 4: Find the volume under z = 6 - x - 2y in the first quadrant.

Answer: 9

Quick Reference

Work through the examples until the process feels automatic. The setup matters more than the arithmetic—get the limits right and the evaluation takes care of itself.