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Examples 5-7
In the previous Examples 3 and 4, the way we specified $D$ and $R$
suggested how to write the integral as a iterated integral.
Sometimes conditions are best interpreted graphically before
deciding on whether to evaluate as Type I or Type II. |
Solution 5: $D$ is enclosed by the straight line $y = x$ and the parabola $y = x^2$ as shown here. To determine the limits of integration we first need to find the points of intersection of $y = x$ and $y = x^2$. These occur when $x^2 = x$. This means $x(x-1)=0$, so $x = 0,\, 1$. Treating $D$ as a Type I region, we fix $x$ between $x=0$ and $x=1$, and integrate with respect to $y$ along the black vertical line, getting the iterated integral $\displaystyle I \ = \ \int_0^1\left(\int_{x^2}^{x}\, (3x + 4y)\, dy\right)\,dx$ DO: Evaluate $I$. |
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Example 6: The region $D$ that
is shown here is Type I but is not Type II: Fixing $x$
and integrating first with respect to $y$ along the vertical
black line makes good sense because then we have the same
curve $f_1(x)$ along the bottom and $f_2(x)$ along the top:
$$ D \ = \ \Bigl\{\,(x,\,y) : f_1(x) \le y \le f_2(x),\ \ a
\le x \le b\,\Bigl\}$$ for suitable choices of $a,\, b$ and
functions $f_1(x),\, f_2(x)$ giving us: $$ \iint_D\,
f(x,\,y)\, dA = \int_a^b \left(\int_{f_1(x)}^{f_2(x)}\,
f(x,\,y)\, dy\right) dx\,.$$But if we had chosen to fix $y$,
then the integral with respect to $x$ would sometimes split
into two parts, as is shown with the red horizontal
lines. This would make evaluate of this integral more
complicated -- for one thing, there would be more than one
integral. |
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Example 7: Similarly, the region $D$ shown here is Type II but not Type I. Fixing $y$ and integrating first with respect to $x$ along the horizontal black line makes good sense because then $$ D \ = \ \Bigl\{\,(x,\,y) : g_1(y) \le x \le g_2(y),\ \ c \le y \le d\,\Bigl\}$$ for suitable choices of $c,\, d$ and functions $g_1(y),\, g_2(y)$. In this case $$ \iint_D\, f(x,\,y)\, dxdy = \int_c^d \left(\int_{g_1(y)}^{g_2(y)}\, f(x,\,y)\, dx\right) dy\,.$$But if we had chosen to fix $x$, then the integral with respect to $y$ would sometimes splits into two parts as shown by the red vertical lines. Again, this would make the integral(s) more complicated. |