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# Double Integrals over Rectangular Regions

## Introduction to Double Integrals

We've handled limits and derivatives in three dimensions; what about integrals? Remember what we know about integrals in two dimensions: the integral of $$f(x)$$ from a to b gives the area under that function between those limits.

When we go from two dimensions to three, area becomes volume, and instead of just having limits on x, we'll also have limits on y, forming a region $$R$$ in the xy plane.

To find the volume of this body, cut it into thin slices in one direction, as shown below.

Then the volume is given by the integral $V=\int A(x)\ dx$ where the area is $A(x)=\int f(x,y)\ dy.$ Therefore, the volume is $V = \iint f(x,y)\ dy\ dx.$

#### Fubini's Theorem

Let $$f$$ be continuous on the rectangular region $$R=\{(x,y)\ :\ a \leq x \leq b, c \leq y \leq d\}.$$ Then $\iint_R f(x,y)\ dA = \int_c^d \int_a^b f(x,y)\ dx\ dy = \int_a^b \int_c^d f(x,y)\ dy\ dx.$ In other words, the order of integration over a rectangular region doesn't matter.

How do we actually evaluate one of these double integrals? The answer is similar to what we found with partial derivatives; when integrating with respect to one variable, treat the other variable as a constant.

Evaluate the following integral.

$\int_1^3 \int_0^1 1+4xy\ dx\ dy$

#### Solution

Start with the inner integral and work outwards. Thus, integrate the function with respect to x first, treating y as a constant:

\begin{align} \int_1^3 \int_0^1 &1+4xy\ dx\ dy\\ &= \int_1^3 \bigg[ x+2x^2y \bigg]_0^1\ dy\\ &= \int_1^3 \bigg[ (1+2y)-(0+0) \bigg]\ dy\\ &= \int_1^3 1+2y\ dy\\ &= y+y^2\bigg|_1^3\\ &= (3+9)-(1+1) = \ans{10} \end{align}

#### Alternate Solution

Note that if we switch the order of integration, we get the same result (Fubini's Theorem):

\begin{align} \int_0^1 \int_1^3 &1+4xy\ dy\ dx\\ &= \int_0^1 \bigg[ y+2xy^2 \bigg]_1^3\ dx\\ &= \int_0^1 2+16x\ dx\\ &= 2x+8x^2\bigg|_0^1\\ &= \ans{10} \end{align}

Find the volume of the solid bounded by the surface $$f(x,y)=4+9x^2y^2$$ over the region $$R=\{(x,y)\ :\ -1 \leq x \leq 1, 0 \leq y \leq 2\}.$$

#### Solution

The volume of this solid is the double integral of the function over the given rectangular region:

\begin{align} V &= \int_{-1}^1 \int_0^2 4+9x^2y^2\ dy\ dx &= \int_{-1}^1 \bigg[ 4y+3x^2y^3 \bigg]_0^2\ dx\\ &= \int_{-1}^1 8+24x^2\ dx\\ &= 8x+8x^3 \bigg|_{-1}^1 = \ans{32} \end{align}

Evaluate the following integral, where $$R=\{(x,y)\ :\ 0 \leq x \leq 1, 0 \leq y \leq \ln 2\}.$$

$\iint_R ye^{xy}\ dA$

#### Solution

Take a second to think before doing the integral. Since this is a rectangular region, we can integrate in either order, so we need to decide which order to use. If we integrate with respect to y first, we'll need to use integration by parts, which isn't impossible, but more difficult than what we get if we integrate with respect to x first.

\begin{align} \int_0^{\ln 2} \int_0^1 &ye^{xy}\ dx\ dy\\ &= \int_0^{\ln 2} \bigg[ e^{xy} \bigg]_0^1\ dy\\ &= \int_0^{\ln 2} e^y-1\ dy\\ &= e^y-y\bigg|_0^{\ln 2}\\ &= 2-\ln 2-1 = \ans{1-\ln 2} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)

## Average Value

Recall from Calculus 1 that the average value of a function over an interval from a to b is $\overline{f} = \dfrac{1}{b-a} \int_a^b f(x)\ dx.$ If we generalize this to three dimensions, the length $$b-a$$ becomes the area of the region $$R$$, and the integral becomes a double integral:

$\ans{\overline{f} = \dfrac{1}{\textrm{area of } R} \iint_R f(x,y)\ dA}$

Find the average value of $$f(x,y)=\sin x\ \sin y$$ over the region $$R=\{(x,y)\ :\ 0 \leq x \leq \pi, 0 \leq y \leq \pi\}.$$

#### Solution

The area of this region is $$\pi^2$$, so the average value is

\begin{align} \overline{f} &= \dfrac{1}{\pi^2} \int_0^\pi \int_0^\pi \sin x\ \sin y\ dx\ dy\\ &= \dfrac{1}{\pi^2} \int_0^\pi \bigg[ -\cos x\ \sin y \bigg]_0^\pi\ dy\\ &= \dfrac{1}{\pi^2} \int_0^\pi 2\sin y\ dy\\ &= \dfrac{1}{\pi^2} \bigg[-2\cos y \bigg]_0^\pi\\ &= \ans{\dfrac{4}{\pi^2}} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)