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Double integrals over a bounded region: the general case.

The definition goes as in  1.1:with the notations of Fig.  3, $S_n= \underset{k}{\sum} f(x_k,y_k) \Delta A_k$.


  
Figure 3: Partitioning of a bounded region.
\begin{figure}
\mbox{\epsfig{file=BoundedGrid.eps,height=6cm} }
\end{figure}

Theorem 7.5 (Fubini)   Let f be a function, continuous on a domain D in the plane. Suppose that D is defined by the inequations $a \leq x \leq b$ and $g_1(x) \leq y \leq g_2(x)$. Then:

\begin{displaymath}\int \int_D f(x,y) \; dA \; = \; \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) dy \; dx.
\end{displaymath}

Example 7.6   Compute the integral $\int \int_D xy^2 \; dA$ over the domain D bounded by the ellipse whose equation is $\frac {x^2}{9}+\frac {y^2}{2.25}=1$.


  
Figure 4: An elliptic region.
\begin{figure}
\mbox{\epsfig{file=ellipse.eps,height=6cm} }
\end{figure}

We have:

\begin{displaymath}\frac {x^2}{9}+\frac {y^2}{2.25}=1
\Longleftrightarrow
y^2= \frac 94 \; \cdot \frac {9-x^2}{9} = \frac 14 (9-x^2).
\end{displaymath}

Thus, the integration domain is described by the inequations:

\begin{displaymath}\begin{cases}
-3 \leq x \leq 3 \\
-\frac 12 \sqrt{9-x^2} \leq y \leq \frac 12 \sqrt{9-x^2}
\end{cases}.
\end{displaymath}

Now we can compute the integral:
\begin{align*}I = \int \int_D xy^2 \; dA
&= \int_{-3}^{3} \int_{-\frac 12 \sqrt{...
...d
&=\frac {1}{60} \left[ (9-x^2)^{5/2} \right]_{-3}^{3}
\quad & =0.
\end{align*}
Note that the last integral could not have been computed by hand, as it is the integral of an odd function over a interval which is symmetric about the origin.


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