THE DOUBLE INTEGRAL OVER A REGION

We start with a closed bounded region $\Omega $ in the $xy$-plane. We assume that $\Omega $ is a basic region; that is, we assume that the boundary of $\Omega $ consists of a finite number of continuous arcs MATH. Now let's suppose that $f$ is some function continuous on $\Omega $. We want to define the double integral
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To do this, we surround $\Omega $ by a rectangle $R.$ We now extend $f$ to all of $R$ by setting $f$ equal to $0$ outside of $\Omega $. This extended function $f$ is bounded on $R$, and it is continuous on all of $R$ except possibly at the boundary of $\Omega $. In spite of these possible discontinuities, $f$ is still integrable on $R$; that is, there still exists a unique number $I$ such that
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This number $I$ is by definition the double integral
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We define the double integral over $\Omega $ by setting
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If $f$ is nonnegative over $\Omega $, the extended $f$ is nonnegative on all of $R$. The double integral gives the volume of the solid trapped between the surface $z=f(x,y)$ and the rectangle $R$. But since the surface has height $0$ outside of $\Omega $, the volume outside of $\Omega $ is $0$. It follows then that
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gives the volume of the solid $T$ bounded above by $z=f(x,y)$ and below by $\Omega $:

volume of $T$ =MATH

The double integral
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gives the volume of a solid of constant height $1$ over $\Omega $. In square units this is the area of MATH:
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Below we list four elementary properties of the double integral. They are all analogous to what was in the one-variable case. The MATH referred to is a basic region. The functions $f$ and $g$ are assumed to be continuous on $\Omega $.

I. The double integral is linear:
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II. The double integral preserves order:


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III. The double integral is additive:

if $\Omega $ is broken up into a finite number of basic regions MATH, then
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IV. The double integral satisfies a mean-value condition; namely, there is a point $(x_{0},y_{0})$ for which
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We call $f(x_0,y_0)$ the average value of $f$ on $\Omega .$

The notion of average given in (IV) enables us to write
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This is a powerful intuitive way of viewing the double integral. We will capitalize on it as we go on.

THEOREM MEAN-VALUE THEOREM FOR DOUBLE INTEGRALS.

Let $f$ and $g$ be functions continuous on a basic region $\Omega $. If $g$ is nonnegative on $\Omega $, then there exists a point $(x_0,y_0)$ in MATH for which
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We call $f(x_0,y_0)$ the $g$-weighted average of $f$ on $\Omega .$

Proof. Since $f$ is continuous on MATH and MATH is closed and bounded, we know that $f$ takes on a minimum value $m$ and a maximum value $M $. Since $g$ is nonnegative on $\Omega ,$


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Therefore
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and
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We know that MATH If MATH, then from the last equation we have MATH and the theorem holds for all choices of $(x_0,y_0)$ in $\Omega $. If MATH, then
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and by the intermediate-value theorem there exists $(x_0,y_0)$ in MATH for which
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Obviously then
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THE EVALUATION OF DOUBLE INTEGRALS BY REPEATED INTEGRALS

The Reduction Formulas

If an integral
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proves difficult to evaluate, it is not because of the interval $[a,b]$ but because of the integrand $f$. Difficulty in evaluating a double integral
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can come from two sources: from the integrand $f$ and from the base region $\Omega $. Even such a simple looking integral as MATH is difficult to evaluate if $\Omega $ is complicated.

In this section we introduce a technique for evaluating double integrals of continuous functions over $\Omega $'s that have a simple structure. The fundamental idea of this section is that double integrals over sets of this structure can be reduced to a pair of ordinary integrals.

Type I The projection of $\Omega $ onto the $x$-axis is a closed interval $[a,b]$ and $\Omega $ consists of all points $(x,y)$ with
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Then
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Here we first calculate
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by integrating $f(x,y)$ with respect to $y$ from $y=\phi _1(x)$ to $y=\phi _2(x)$. The resulting expression is a function of $x$ alone, which we then integrate with respect to $x$ from $x=a$ to $x=b$.

Type II. The projection of $\Omega $ onto the $y$-axis is a closed interval $[a,b]$ and $\Omega $ consists of all points $(x,y)$ with
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Then
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This time we first calculate
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by integrating $f(x,y)$ with respect to $x$ from $x=\phi _1(y)$ to $x=\phi _2(y)$. The resulting expression is a function of $y$ alone, which we then integrate with respect to $y$ from $y=a$ to $y=b$. These formulas are easy to understand geometrically.

The Reduction Formulas Viewed Geometrically

Let's take $f$ as nonnegative and $\Omega $ of type I. The double integral over $\Omega $ gives the volume of the solid $T$ that is trapped between $\Omega $ and the surface $z=f(x,y):$
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We can also compute the volume of $T$ by the method of parallel cross sections. Let $A(x)$ be the area of that cross section of $T$ that has first coordinate $x$. Then
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Since
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we have
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Combining the two equations, we have the first reduction formula
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The other reduction formula can be obtained in a similar manner.

Computations

Problem 1. Evaluate
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with $\Omega $ the region bounded by
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Solution. By projecting $\Omega $ onto the $x$-axis we obtain the interval $[-1,1].$ The region $\Omega $ consists of all points $(x,y)$ with
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This is a region of type I.
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Problem 2. Evaluate
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with $\Omega $ bounded by
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Solution. By projecting $\Omega $ onto the $y$-axis we obtain the interval $[0,1]$. The region $\Omega $ consists of all points $(x,y)$ with

$0\le y\le 1$ and $-1\le x\le y.$

This is a region of type II.
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We can also project $\Omega $ onto the $x$-axis and express $\Omega $ as a region of type I, but then the lower boundary is defined piecewise and the calculations are somewhat more complicated: setting
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we have $\Omega $ as the set of all points $(x,y)$ with
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thus
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Repeated integrals
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can be written in more compact form by omitting the large parentheses. From now on we will simply write
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Problem 3. Evaluate
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with $\Omega $ the region bounded by
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Solution. The projection of $\Omega $ onto the $x$-axis is the closed interval $[0,1]$, and $\Omega $ can be characterized as the set of all $(x,y) $ with
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Thus
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We can also integrate in the other order. The projection of $\Omega $ onto the $y$-axis is the closed interval $[0,l]$, and $\Omega $ can be characterized as the set of all $(x,y)$ with
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This gives the same result:
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Problem 4. Calculate by double integration the area of the region $\Omega $ that lies between
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Solution. The area of $\Omega $ is given by the double integral
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Here again we can integrate in either order. We can project $\Omega $ onto the $x$- axis and write the boundaries as functions of $x$:
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$\Omega $ is then the set of all $(x,y)$ with
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This gives
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We can also project $\Omega $ onto the $y$-axis and write the boundaries as functions of $y$:
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$\Omega $ is then the set of all $(x,y)$ with
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This gives
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which is also $\frac{a^2}3.$

Symmetry in Double Integration

First we go back to the one-variable case. Let's suppose that $g$ is continuous on an interval that is symmetric about the origin, say MATH

If $g$ is odd, then MATH.

If $g$ is even, then MATH

We have similar results for double integrals.

Suppose that $\Omega $ is symmetric about the $y$-axis .

If $f$ is odd in $x$ if $f(-x,y)=-f(x,y)$ , then MATH

If $f$ is even in $x$ if $f(-x,y)=f(x,y)$ , then MATH

Suppose that $\Omega $ is symmetric about the $x$-axis.

If $f$ is odd in $y$ if $f(x,-y)=-f(x,y)$ , then MATH

If $f$ is even in $y$ if $f(x,-y)=f(x,y)$ , then MATH

As an example take the region bounded by
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Suppose we wanted to calculate
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Symmetry about the $y$-axis gives
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Symmetry about the $x$-axis gives
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Therefore
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As another example we go to Problem 1 and reevaluate
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this time capitalizing on the symmetry of $\Omega $. Note that
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Therefore
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Problem 5. Calculate the volume within the cylinder
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between the planes
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and
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given that
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Solution. The solid in question is bounded below by the disc
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and above by the plane


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The volume is given by the double integral
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Since $\Omega $ is symmetric about the $x$-axis,
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Thus
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Concluding Remarks

When two orders of integration are possible, one order may be easy to carry out, while the other may present serious difficulties. Take as an example the double integral
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with $\Omega $ bounded by
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Projection onto the $x$-axis leads to
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Projection onto the $y$-axis leads to
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The first expression is easy to calculate, but the second is not. The only practical way to handle the second expression is to expand $\cos \frac 12\pi x$ as a series in $x$ and then integrate term by term.

Finally, if $\Omega $, the region of integration, is neither of type I nor of type II, it is usually possible to break it up into a finite number of regions MATH, each of type I or type II. Since the double integral is additive,
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Each of the integrals on the left can be evaluated by the methods of this section.

Exercise

  1. Evaluate the integrals

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  2. Write an equivalent double integral with the order of integration reversed.

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  3. Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola $y=4-x^2$ and the line $y=3x$, while the top of the solid is bounded by the plane MATH

  4. Find the volume of the solid in the first octant bounded by the coordinate planes, the plane $x=3,$ and the parabolic cylinder MATH

  5. Find the volume of the region that lies under the paraboloid $z=x^{2}+y^{2}$ and above the triangle enclosed by the lines $y=x,x=0,$ and $x+y=2$ on the xy-plane.MATH

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