4.1 Double Integrals over Rectangles

Understanding Double Integrals over Rectangles

Double integrals extend single-variable integration to two dimensions, letting you calculate the volume beneath a surface $z = f(x,y)$ over a flat rectangular region. They're the foundation for all the multiple integral techniques you'll encounter in this unit.

The core strategy is iterated integration: instead of tackling both variables at once, you integrate with respect to one variable while treating the other as a constant, then integrate the result with respect to the second variable. Fubini's Theorem is what makes this legal, and it also lets you swap the order of integration when one order is easier than the other.

Evaluation of Double Integrals

The notation $\iint_R f(x,y)\, dA$ means "integrate the function $f(x,y)$ over the rectangular region $R$." For a rectangle $R = [a,b] \times [c,d]$, this becomes an iterated integral:

$\iint_R f(x,y)\, dA = \int_a^b \int_c^d f(x,y)\, dy\, dx$

The outer integral's limits correspond to the outer variable (here $x$), and the inner integral's limits correspond to the inner variable (here $y$).

How to evaluate step by step:

1. Start with the inner integral. Integrate $f(x,y)$ with respect to $y$, treating $x$ as a constant. Evaluate from $y = c$ to $y = d$. The result is a function of $x$ alone.
2. Take that result and integrate it with respect to $x$ from $a$ to $b$. This gives you a single number.

Fubini's Theorem states that for any continuous function $f$ on a rectangle $R$:

$\int_a^b \int_c^d f(x,y)\, dy\, dx = \int_c^d \int_a^b f(x,y)\, dx\, dy$

Both orders produce the same answer. This is only guaranteed on rectangles (and more generally when $f$ is continuous on the region), but for this section that's always the case.

Volume Calculation with Double Integrals

Geometrically, $\iint_R f(x,y)\, dA$ gives the volume of the solid that sits above the rectangle $R$ in the $xy$-plane and below the surface $z = f(x,y)$ (assuming $f(x,y) \geq 0$ on $R$).

Setting up a volume problem:

1. Identify the surface $z = f(x,y)$. This could be a paraboloid like $z = x^2 + y^2$, a plane like $z = 3x + 2y$, or any other function of two variables.
2. Determine the rectangular region $R$. For example, $R = [0, 2] \times [1, 3]$ means $x$ ranges from 0 to 2 and $y$ ranges from 1 to 3.
3. Write the volume as $V = \iint_R f(x,y)\, dA$ and evaluate using iterated integration.

Quick example: Find the volume under $z = 6 - 2x$ over $R = [0,1] \times [0,2]$.

$V = \int_0^1 \int_0^2 (6 - 2x)\, dy\, dx = \int_0^1 \left[(6-2x)y\right]_0^2 dx = \int_0^1 2(6-2x)\, dx = \int_0^1 (12 - 4x)\, dx$

$= \left[12x - 2x^2\right]_0^1 = 12 - 2 = 10$

Average Value over Rectangular Regions

The average value of $f(x,y)$ over a rectangle $R$ works just like the 1D average value formula, but now you divide by the area of the region instead of the length of an interval:

$f_{\text{avg}} = \frac{1}{A} \iint_R f(x,y)\, dA$

where $A = (b - a)(d - c)$ is the area of the rectangle $[a,b] \times [c,d]$.

Steps:

1. Compute the area $A$ of the rectangle.
2. Evaluate $\iint_R f(x,y)\, dA$ using iterated integration.
3. Divide the integral's value by $A$.

For the example above, the average value of $f(x,y) = 6 - 2x$ over $[0,1] \times [0,2]$ would be $\frac{10}{(1)(2)} = 5$.

Order of Integration in Rectangles

On rectangular regions, you always have the freedom to integrate in either order. The question is which order makes the computation simpler.

Choosing the better order:

• Look at the integrand and ask: is it easier to integrate with respect to $x$ first, or $y$ first?
• For something like $f(x,y) = x\, e^{xy}$, integrating with respect to $x$ first is messy, but integrating with respect to $y$ first gives a straightforward exponential integral.
• Polynomials are usually equally easy in either order. The payoff of switching order shows up most with products of functions, exponentials, or trigonometric expressions where one variable creates a simpler antiderivative.

A good habit: before computing, glance at both orders. Spending 30 seconds choosing the right order can save you minutes of messy algebra.