9.2 Evaluation of double integrals over rectangles

Iterated Integrals

Double integrals over rectangles let you calculate quantities like area and volume by breaking a two-variable problem into two one-variable integrals. You evaluate them one at a time: compute the inner integral first, then feed that result into the outer integral.

Definition and Notation

An iterated integral is written as:

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

You can also write it with explicit parentheses to make the evaluation order clearer:

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

Here's how to evaluate one:

1. Inner integral first. Integrate $f(x,y)$ with respect to $y$ from $c$ to $d$, treating $x$ as a constant. This produces a function of $x$ alone.
2. Outer integral second. Integrate that result with respect to $x$ from $a$ to $b$. This gives you a number.

Quick example: Evaluate $\int_0^1 \int_0^2 xy^2\, dy\, dx$.

1. Inner integral (with respect to $y$, treating $x$ as constant):

$\int_0^2 xy^2\, dy = x \cdot \frac{y^3}{3}\Big|_0^2 = x \cdot \frac{8}{3} = \frac{8x}{3}$

1. Outer integral (with respect to $x$):

$\int_0^1 \frac{8x}{3}\, dx = \frac{8}{3} \cdot \frac{x^2}{2}\Big|_0^1 = \frac{4}{3}$

Types of Limits of Integration

• Constant limits are fixed values that don't depend on the other variable. The example above, $\int_0^1 \int_0^2 xy^2\, dy\, dx$, has all constant limits. This is the standard setup for rectangular regions.
• Variable limits depend on the outer variable. For instance, $\int_0^1 \int_0^x y^2\, dy\, dx$ has an upper limit of $x$ on the inner integral, meaning the range of $y$ changes as $x$ changes.
• Variable limits show up when the region of integration is not a rectangle. For this unit (rectangular regions), you'll mostly work with constant limits on both integrals.

Reversing Order of Integration

Swapping the Order of Integration

For rectangular regions with constant limits, Fubini's Theorem guarantees that you can swap the order of integration freely, as long as $f(x,y)$ is continuous on the rectangle:

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

This is useful when one order leads to a much simpler antiderivative than the other. When you reverse the order:

• The old inner limits become the new outer limits (and vice versa).
• The differentials swap: $dy\, dx$ becomes $dx\, dy$.

For non-rectangular regions (variable limits), reversing the order requires you to re-derive the limits from the geometry of the region, which is more involved.

Visualizing the Change of Order

Think of the two orders as slicing the region in different directions:

• $dy\, dx$ order: You slice the region into thin vertical strips. For each fixed $x$, you integrate along $y$ (bottom to top), then sweep those strips from left to right.
• $dx\, dy$ order: You slice into thin horizontal strips. For each fixed $y$, you integrate along $x$ (left to right), then sweep those strips from bottom to top.

Both approaches cover the same region and yield the same answer. The choice comes down to which integrand is easier to handle in a given order.

Applications of Double Integrals

Area Calculation

To find the area of a planar region $R$, integrate the constant function 1 over that region:

$\text{Area} = \iint_R 1\, dA$

Example: Find the area of the region bounded by $y = x^2$ and $y = 4$.

The curves intersect where $x^2 = 4$, so $x = -2$ and $x = 2$. Using symmetry (or just integrating from $-2$ to $2$), set up:

$\int_{-2}^{2} \int_{x^2}^{4} dy\, dx$

The inner integral gives $4 - x^2$, and then $\int_{-2}^{2}(4 - x^2)\, dx = \frac{32}{3}$.

Note: this particular example uses variable limits, so it goes beyond a purely rectangular region. On a rectangle like $[a,b] \times [c,d]$, the area is simply $(b-a)(d-c)$.

Volume Calculation

The most common application over rectangles is computing volume. If a surface $z = f(x,y) \geq 0$ sits above a rectangular base $R = [a,b] \times [c,d]$, the volume of the solid between the surface and the $xy$-plane is:

$V = \iint_R f(x,y)\, dA$

Here $f(x,y)$ gives the height of the surface above each point $(x,y)$ in the base.

Example: Find the volume under $z = xy$ over the rectangle $[0,2] \times [0,1]$.

1. Set up the integral: $\int_0^2 \int_0^1 xy\, dy\, dx$
2. Inner integral: $\int_0^1 xy\, dy = x \cdot \frac{y^2}{2}\Big|_0^1 = \frac{x}{2}$
3. Outer integral: $\int_0^2 \frac{x}{2}\, dx = \frac{x^2}{4}\Big|_0^2 = 1$

The volume is $1$ cubic unit. You can verify this by reversing the order of integration and confirming you get the same result.