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Double integrals over a rectangle.

Let f(x,y) be a function defined on a rectangular domain $R=\{ (x,y)
\vert a \leq x \leq b, \; c \leq y \leq d \}$. We divide the rectangle Rwith a network of lines parallel to the coordinate axes, and we number the little rectangles A1, A2, and so on. On each little rectangle Ak we choose a point (xk, yk), as on Fig  1. The area of such a little rectangle is denoted $\Delta A = \Delta x \Delta y$.
  
Figure 1: Partitioning of a rectangular region.
\begin{figure}
\mbox{\epsfig{file=RectangularGrid.eps,height=6cm} }
\end{figure}

Define $S_n= \underset{k}{\sum} f(x_k,y_k) \Delta A_k$. If this sum has a finite limit when we refine indefinitely the network, then this limit is called the double integral of f over R and is denoted

\begin{displaymath}\int \int_R f(x,y) dA
\end{displaymath}

Proposition 7.1         
1.
$\int \int_R [ f(x,y) + g(x,y) ] dA =
\int \int_R f(x,y) dA + \int \int_R g(x,y) dA$.
2.
$\int \int_R \alpha f(x,y) dA = \alpha \int \int_R f(x,y) dA $.
3.
If $\forall (x,y) \in R, \; f(x,y) \geq 0$, then $\int \int_R f(x,y) dA \geq 0$.
4.
If $\forall (x,y) \in R, \; f(x,y) \leq g(x,y)$, then $\int \int_R f(x,y) dA \leq \int \int_R g(x,y) dA$.

Proposition 7.2   Suppose that the rectangle R is the union of two rectangles R1 and R2, as dispayed on Fig.  2, then:

\begin{displaymath}\int \int_R \alpha f(x,y) dA = \int \int_{R_1} \alpha f(x,y) dA + \int \int_{R_2} \alpha f(x,y) dA
\end{displaymath}


  
Figure 2: The union of two rectangles.
\begin{figure}
\mbox{\epsfig{file=2rectangles.eps,height=6cm} }
\end{figure}

Such a double integral can be viewed as the volume of the three-dimensional region enclosed by the domain R in the xy-plane, the graph of f and the planes parallel to the coordinates axes and perpendicular to the sides of R.

Theorem 7.3 (Fubini)   If f is continuous in a region containing the rectangle R, then:

\begin{displaymath}\int \int_R \alpha f(x,y) dA = \int_{x=a}^{x=b} \int_{y=c}^{y...
...dx
= \int_{y=c}^{y=d} \int_{x=a}^{x=b} \alpha f(x,y) dx \; dy
\end{displaymath}

Example 7.4   f(x,y)=xy+1 over $R=\{ (x,y) \vert 1 \leq x \leq 3 ; \; 1 \leq y \leq 4 \}$.


\begin{align*}I & = \int_{y=1}^{y=4} \int_{x=1}^{x=3} (xy+1) \; dx \; dy \\
\qu...
...=4} (4y +2 )dy \\
\quad &= \left[ 2y^2+2y \right]_1^4 = 40 -4 =36.
\end{align*}


next up previous contents
Next: Double integrals over a Up: Double integrals. Previous: Double integrals.
Noah Dana-Picard
2001-05-30