Double Integration- How to Do Average of Double Integration
What Is the Average of a Double Integral?
When you integrate a function over a 2D region, you get a single number. That number represents the total accumulated value across the region. But what if you want to know the average value of the function over that region?
The average value of a function f(x,y) over a region R tells you what the function would equal if it were spread out uniformly across that entire area. It's the 2D equivalent of finding the mean of a dataset.
The formula is straightforward:
Average value = (1/Area of R) × ∬f(x,y) dA
You take the double integral, then divide by the area of the region. That's it. No tricks.
Why This Formula Works
Think of it this way. In 1D, the average value of f(x) on [a,b] is:
Average = (1/(b-a)) × ∫f(x) dx
Double integrals follow the same logic. The region R has some area A. Dividing the total integral by A "normalizes" it—scales it down so you're looking at a per-unit-area basis.
If f represents height above a surface, the double integral gives you the volume under that surface. Dividing by the base area gives you the average height. Simple.
How to Calculate the Average of a Double Integral
Step 1: Set Up Your Double Integral
Identify your function f(x,y) and your region R. Write out the double integral with appropriate bounds.
Step 2: Evaluate the Double Integral
Integrate with respect to one variable first, treating the other as constant. Then integrate the result with respect to the second variable.
Step 3: Find the Area of Region R
If R is described by simple bounds, you already have this from Step 1. If not, you may need to set up a separate integral for the area.
Step 4: Divide
Take your result from Step 2 and divide by the area from Step 3.
Example: Average Temperature on a Rectangular Plate
Suppose a rectangular plate occupies 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3. The temperature at any point is given by T(x,y) = 20 + 5x + 3y. What is the average temperature?
Step 1: The double integral is ∫₀³ ∫₀² (20 + 5x + 3y) dx dy
Step 2: Inner integral with respect to x:
∫₀² (20 + 5x + 3y) dx = [20x + (5/2)x² + 3yx]₀² = 40 + 10 + 6y = 50 + 6y
Outer integral with respect to y:
∫₀³ (50 + 6y) dy = [50y + 3y²]₀³ = 150 + 27 = 177
Step 3: Area of rectangle = 2 × 3 = 6
Step 4: Average temperature = 177 / 6 = 29.5
Units would be degrees (whatever your temperature scale is).
Example: Average Value Over a Circular Region
Find the average value of f(x,y) = x² + y² over the unit disk (x² + y² ≤ 1).
Switch to polar coordinates. The double integral becomes:
∬(r²) r dr dθ = ∬ r³ dr dθ
Bounds: 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
Integrate with respect to r: ∫₀¹ r³ dr = 1/4
Integrate with respect to θ: ∫₀²π (1/4) dθ = (1/4)(2π) = π/2
Area of unit disk = π(1)² = π
Average value = (π/2) / π = 1/2
Common Mistakes to Avoid
- Forgetting to divide by area. This is the most common error. The double integral alone gives you the total, not the average.
- Using the wrong area. Make sure you're calculating the actual geometric area of region R, not just copying bounds from your integral.
- Assuming symmetry where there isn't any. Just because a region looks symmetric doesn't mean the function is symmetric over it.
- Messy bounds in polar coordinates. When converting to polar, remember the Jacobian adds an r. Your bounds must fully describe the region.
Average vs. Total: When to Use Which
| What You Want | Formula | Use Case |
|---|---|---|
| Total accumulated value | ∬f(x,y) dA | Total volume, total mass, total quantity |
| Average value per unit area | (1/A) × ∬f(x,y) dA | Average temperature, mean density, average rate |
| Weighted average | (1/A) × ∬w(x,y)·f(x,y) dA | When you need to weight by some function |
Practical Applications
Engineers use average values over surfaces to calculate mean stress, mean heat flux, or average pressure on a contact area.
Statisticians encounter 2D average values when working with joint probability distributions—the average of a probability density function over a region gives you the probability of that region.
In physics, finding the center of mass of a lamina requires average values. The x-coordinate of the center of mass is the average of x weighted by density over the region.
Quick Reference: The Process in 4 Steps
- Integrate f(x,y) over region R to get total T
- Calculate the area A of region R
- Divide: Average = T ÷ A
- Check units: should be f(x,y) units per square unit of area
The Bottom Line
Finding the average of a double integral is elementary once you understand what you're doing. Integrate the function. Find the area. Divide. The math takes about 30 seconds once you've set it up correctly.
The hard part is setting up the integral with correct bounds. If you can do that, the rest is arithmetic.