This section has three parts. In Part 1 we show how to find
given its gradient

In Part 2 we show that, although all gradients
are expressions of the form

set
and
,
not all such expressions are gradients. In Part 3 we tackle the problem of
recognizing which expressions
are actually gradients.

*Problem1.* Find
given that

*Solution.* Since

we
have

Integrating
with respect to
,
we have

where
is independent of
but may depend on
.
Differentiation with respect to
gives

The
two equations for
can be reconciled only by having

This
means that

We could have started by integrating with respect to and then differentiating with respect to . This procedure would have yielded the same result.

**Problem 2.** Find
given that

*Solution.* Here we have

We will proceed as in the first problem. Integrating
with respect to
,
we have

with
independent of
.
Differentiation with respect to
gives

The two equations for
can be reconciled only by having

This means that

The function

is the **general solution** of the vector differential equation

Each particular solution can be obtained by assigning a particular value to
the constant of integration
.

The next problem shows that not all linear combinations are gradients.

**Problem 3.** Show that
is not a gradient.

*Solution.* Suppose on the contrary that it is a gradient. Then there
exists a function
such that

Obviously

and thus

This contradicts the mixed partial derivative theorem: the four partial
derivatives under consideration are everywhere continuous and thus, according
to the mixed partial derivative theorem, we must have

This contradiction shows that
is not a gradient.

We come now to the problem of recognizing which linear combinations

are actually gradients. But first we need to review some ideas and establish
some new terminology.

An open set (in the plane or in three-space) is said to be *connected*
if any two points of the set can be joined by a polygonal path that lies
entirely within the set. An open connected set is called an *open
region*. A curve

is said to be **closed** if it begins and ends at the same point:

It is said to be **simple** if it does not intersect itself:

As is intuitively clear, a simple closed curve in the plane separates the
plane into two disjoint open connected sets: a bounded inner region consisting
of all points surrounded by the curve and an unbounded outer region consisting
of all points not surrounded by the curve. (**Jordan Curve
Theorem**.)

Finally, we come to the notion we need in our work with gradients:

Let
be an open region of the plane, and let
be an arbitrary simple closed curve.
is said to be **simply connected** if
is in
implies the inner region of
is in

**THEOREM** Let
and
be functions of two variables, each continuously differentiable on a simply
connected open region
.
The linear combination
is a gradient on
if and only if

for all
.

Proof. A complete proof of this theorem is given in an advanced calculus
course. We will prove the result under the additional assumption that
has the form of an open rectangle with sides parallel to the coordinate axes.
Suppose that
is a gradient on this open rectangle
,
say

Since

we have

Since
and
have continuous first partials,
has continuous second partials. Thus, according to the mixed partials theorem
we have

Conversely, suppose that

We must show that
is a gradient on
.
To do this, we choose a point
from
and form the function

The first integral is independent of
.
Hence it follows from the fundamental theorem of calculus that

Differentiating
with respect to
we have

The first term is
since once again we are differentiating with respect to the upper limit. In
the second term the variable
appears under the sign of integration. It can be shown that, since
and
are continuous,

Anticipating this result, we have

We have now shown that

It follows that
is the gradient of
on

**Example 4**. The vector functions

and

are both everywhere defined. The first vector function is the gradient of a
function everywhere defined:

The second is not a gradient:

and so

**Example 5.** The vector function defined on the *punctured
disc*
by setting

satisfying

We will see later that
is not a gradient on that set. The set is not simply connected so the theorem
does not apply.

Let
be a function continuous on the rectangle

Our object is to define the double integral

Recall that to define the integral

we first introduced some auxiliary notions: partition
,
upper sum
,
and lower sum
We were then able to define

as the unique number
that satisfies the inequality

We will follow exactly the same procedure to define the double integral

First we explain what we mean by a *partition of the rectangle*
.
To do this, we begin with a partition

and a partition

The set

is called a **partition** of
.
consists of all the grid points

The partition
breaks up
into
nonoverlapping rectangles
where for each fixed

On each rectangle
,
the function
takes on a maximum value
,
and a minimum value
The sum of all the products

is called the
**upper sum** for
:

The sum of all the products

is called the
**lower sum** for
:

**Example 1. **Consider the function

on the rectangle

As a partition of
take

and as a partition of
take

The partition
then breaks up the initial rectangle into the six rectangles.

On each rectangle
,
the function
takes on its maximum value
at the point
,
the corner farthest from the origin:

Thus

(area of ) +(area of ) + (area of )

+
(area
of
)
+
(area of
) +
(area
of
)

On each rectangle
,
takes on its minimum value
at the point
,
the corner closest to the origin:

Thus

(area of ) +(area of ) + (area of )

+
(area
of
)
+
(area of
) +
(area
of
)

We return now to the general situation. As in the one-variable case, it can be
shown that, if
is continuous, then there exists one and only one number
that satisfies the inequality

**Definition of the double integral over a rectangle
**

The unique number
that satisfies the inequality

is called the **double integral** of
over
and is denoted by

If
is nonnegative on the rectangle
,
the equation

represents a surface that lies above
.
In this case the double integral

gives the volume of the solid that is bounded below by
and bounded above by the surface

To see this, consider a partition of . breaks up into subrectangles ; and thus the entire solid into parts . Since contains a rectangular solid with base and height , we must have

(area of ) volume of .

Since is contained in a rectangular solid with base and height , we must have

volume of (area of ).

In short, for each pair of indices and , we must have

(area of ) volume of (area of ).

Adding up these inequalities, we are forced to conclude that

volume of .

Since is arbitrary, the volume of must be the double integral:

volume of =

The double integral

gives the volume of a solid of constant height
erected over
.
In square units this is just the area of
:

Double integrals are generally computed by techniques that we will take up later. It is possible, however, to evaluate simple double integrals directly from the definition.

**Problem 2.** Evaluate

where
is the rectangle

*Solution.* Here

We begin with a partition

and a partition

This gives

as an arbitrary partition of
On each rectangle
,
has constant value
.
This gives
and
throughout. Thus

Similarly

Since

and
was chosen arbitrarily, the inequality must hold for all partitions
of
.
This means that

Remark. If

gives the volume of the rectangular solid of constant height
erected over the rectangle
.

**Problem 3.** Evaluate

where

Solution. With with a partition

and a partition

we have

as an arbitrary partition of
On each rectangle
the function

has a maximum

and a minimum

Thus

and

For each pair of indices
and
,

This means that for arbitrary
we have

The middle term can be written

The first double sum reduces to

The second double sum reduces to

The sum of these two numbers

satisfies the inequality

The integral is therefore
:

Remark. This last integral should not be interpreted as a volume. The expression does not keep a constant sign on

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