Recall our definition of the definite integral of a function of a single variable:

$\qquad$

Let $f(x)$ be defined on $[a,b]$ and let $x_{0},x_{1},\ldots,x_{n}$ be a partition of $[a,b]$. For each $[x_{i-1},x_{i}]$, let $x_{i}^{*} \in [x_{i-1},x_{i}]$. Then $$ \int_{a}^{b}f(x)\,dx = \lim_{\max \Delta x_{i} \rightarrow 0} \sum_{i=1}^{n}f(x_{i}^{*})\Delta x_{i}. $$

Take a quick look at the Riemann Sum Tutorial.

We can extend this definition to define the integral of a function of two or more variables.

Double Integral of a Function of Two Variables

Let $f(x,y)$ be defined on a closed and bounded region $R$ of the $xy$-plane. Set up a grid of vertical and horizontal lines in the $xy$-plane to form an inner partition of $R$ into $n$ rectangular subregions $R_{k}$ of area $\Delta A_{k}$, each of which lies entirely in $R$. (Ignore the rectangles that are not entirely contained in $R$) Choose a point $(x_{k}^{*}, y_{k}^{*})$ in each subregion $R_{k}$. The sum $$ \sum_{k=1}^{n} f(x_{k}^{*}, y_{k}^{*}) \Delta A_{k} $$ is called a Riemann Sum. In the limit as we make our grid more and more dense, we define the double integral of $f(x,y)$ over $R$ as $$ \iint\limits_R f(x,y\,)dA= \lim_{\max \Delta A_{k} \rightarrow 0} \sum_{k=1}^{n} f(x_{k}^{*}, y_{k}^{*}) \Delta A_{k} . $$

• If this limit exists, we say that $f$ is integrable over the region of integration $R$.
  
  
• If $f$ is continuous on $R$, then $f$ is integrable over $R$.

Geometric Interpretation of the Double Integral

Notice that as we increase the density of our grid, the sum $\sum\limits_{k=1}^{n}A_{k}$ of the individual rectangles better and better approximates the area of region $R$. In the limit as $\Delta A_{k} \rightarrow 0$, we have $$ \mbox{Area of } R = \iint\limits_{R} dA. $$ Suppose now that $f(x,y) \ge 0$ on $R$. Then $f(x_{k}^{*}, y_{k}^{*})\Delta A_{k}$ is the volume of a rectangular parallelopiped of height $f(x_{k}^{*},y_{k}^{*})$ and base area $\Delta A_k$. Adding up these volumes, we get an appoximation for the volume of the solid above $R$ and below the suface $z=f(x,y)$. Thus, in the limit as $\Delta A_{k} \rightarrow 0$,

Volume of solid above $R$ and below the surface $z=f(x,y)$

$$ \quad = \quad $$

$$ \iint\limits_{R} f(x,y) dA \quad (\mbox{for } \textstyle f(x,y) \ge 0 \mbox{ on } R). $$

We next turn to the actual evaluation of double integrals.

Iterated Integrals

In the double integral $\iint\limits_{R} f(x,y) \,dA$, the differential $dA$ may be viewed informally as an infinitesimal area of a rectangle inside $R$ with dimensions $dy$ and $dx$. For the kinds of "ordinary" functions and regions we'll be concerned with, \begin{eqnarray*} \iint\limits_{R} f(x,y)dA & = & \int_{a}^{b}\left[ \int_{g_{1}(x)}^{g_{2}(x)} f(x,y\,)dy \right] \,dx = \int_{a}^{b}\int_{g_{1}(x)}^{g_{2}(x)} f(x,y\,)dy\,dx \\ & = & \int_{c}^{d} \left[ \int_{h_{1}(y)}^{h_{2}(y)} f(x,y) \,dx \right] \,dy = \int_{c}^{d}\int_{h_{1}(y)}^{h_{2}(y)} f(x,y)\, dx\,dy \end{eqnarray*}

where the limits of integration are determined by the region $R$ over which we are integrating.

• These integrals are called iterated integrals, since we integrate more than once.
  
  
• We integrate "from the inside out." That is, in $\displaystyle\int_{a}^{b}\int_{g_{1}(x)}^{g_{2}(x)} f(x,y)\,dy\,dx$, we first integrate $f(x,y)$ with respect to $y$ and evaluate it at $g_{2}(x)$ and $g_{1}(x)$. We then integrate the result with respect to $x$ and evaluate the outcome at $a$ and $b$.
  
  
• Iterated triple integrals $\displaystyle\iiint\limits_{G} f(x,y,z)dV$ can be defined in a similar way.

An example will make these ideas more concrete.

Example

Let's evaluate the double integral $\displaystyle \iint\limits_{R} 6xy\, dA$, where $R$ is the region bounded by $y=0$, $x=2$, and $y=x^{2}$. We will verify here that the order of integration is unimportant:

Integrating first with respect to $y$, then with respect to $x$: \begin{eqnarray*} \iint\limits_{R} 6xy dA & = & \int_{0}^{2}\int_{0}^{x^{2}} 6xy\, dy\,dx \\ & = & \int_{0}^{2}\left[ \left. 3xy^{2} \right|_{y=0}^{x^{2}} \right]\, dx \\ & = & \int_{0}^{2} 3x^{5}\, dx\\ & = & \left. \frac{1}{2} x^{6} \right|_{x=0}^{2}\\ & = & \frac{1}{2}(64)-\frac{1}{2}(0)\\ & = & 32 \end{eqnarray*}

Integrating first with respect to $x$, then with respect to $y$: \begin{eqnarray*} \iint\limits_{R} 6xy dA & = & \int_{0}^{4}\int_{\sqrt{y}}^{2} 6xy \,dx\,dy \\ & = & \int_{0}^{4}\left[ \left. 3x^{2y} \right|_{x=\sqrt{y}}^{2} \right] \, dy \\ & = & \int_{0}^{4}\left( 12y-3y^{2} \right) \, dy \\ & = & \left. \left( 6y^{2} -y^{3} \right) \right|^{4}_{y=0} \\ & = & \left( 6(4)^{2}-(4)^{3} \right) - \left( 6(0)^{2}- (0)^{3} \right) \\ & = & 32 \end{eqnarray*}

so $\displaystyle \iint\limits_{R} 6xy \, dA =32$ here, regardless of the order in which we carry out the integration, as long as we are careful to set up the limits of integration correctly.

Now for a triple integral...

Example

We will evaluate the triple integral $\displaystyle \int_{0}^{2}\int_{-1}^{y^{2}}\int_{1}^{z} yz \, dx \,dz\,dy.$

\begin{eqnarray*} \int_{0}^{2}\int_{-1}^{y^{2}}\int_{1}^{z} yz \, dx \,dz\,dy & = & \int_{0}^{2}\int_{-1}^{y^{2}} \left[ \left. (xyz) \right|_{x\,=\,1}^{x\,=\,z} \right]\,dz\,dy \\ & = & \int_{0}^{2}\int_{-1}^{y^{2}} \left( yz^{2}-yz \right)\,dz\,dy \\ & = & \int_{0}^{2} \left[ \left. \left ( \frac{yz^{3}}{3}-\frac{yz^{2}}{2}\right) \right|_{z\,=\,-1}^{z\,=\,y^{2}} \right] \,dy\\ & = & \int_{0}^{2} \left( \frac{y^{7}}{3} - \frac{y^{5}}{2} + \frac{5y}{6} \right)\,dy\\ & = & \left. \left(\frac{y^{8}}{24}-\frac{y^{6}}{12}+\frac{5y^{2}}{12} \right) \right|_{0}^{2}\\ & = & \frac{264}{24}-\frac{64}{12}+\frac{20}{12}\\ & = & \frac{84}{12}\\ & = & 7. \end{eqnarray*}	Integrate with respect to $x$ first.
	Next integrate with respect to $z$.
	Finally, integrate with respect to $y$.

We evaluate $\displaystyle\iint\limits_{R} f(x,y) \, dA$ as an iterated integral: \begin{eqnarray*} \iint\limits_{R} f(x,y) \, dA & = & \int_{a}^{b}\int_{g_{1}(x)}^{g_{2}(x)} f(x,y\,)dy\,dx \\ & = & \int_{c}^{d}\int_{h_{1}(y)}^{h_{2}(y)} f(x,y)\, dx\,dy \end{eqnarray*} for "ordinary" regions $R$ and functions $f(x,y)$.