Consider the 1-form defined by
. This is the form which
matches the proposed solution since
Furthermore, it is the volume element. To see this, choose an
orthonormal basis
where
where
.
Then
and
if and
only if
, as desired.
Let
be an orientation-preserving
-cube, i.e.
. Let
be an orthonormal basis with
the usual orientation.
By Theorem 4-9, we have
and so applying this to
gives
Since
is
orientation preserving, it must be that
.
Now,
Now,
is orientation preserving, so
must
have the usual orientation, i.e.
. But then
. Since this
is also equal to
by orthonormality, it follows that
the value is precisely 1, and so it is the volume element in the sense of
this section.
The generalization is
defined by
where
is the unit outward normal at
. As in the 2-dimensional
case,
if the
are an orthonormal basis
with orientation
(where
was the orientation used to determine
the outward normal). So
is the volume element
.
Expanding in terms of cofactors of the last row gives:
As in the 2-dimensional case,
for
some scalar
and so
for all
. Letting
, we get
One can use singular 2-cubes of the form
. The quantities
,
, and
calculate out to
,
, and
. So the surface area is
.
Apply part (a) with
and
.
One has
and
.
So
.
Although it is not stated, it is assumed that
is orientable.
By Problem 1-7,
is inner product preserving and so it maps orthonormal
bases to orthonormal bases. Further, if
is a singular
-cube which
is a coordinate system for
in a neighborhood of
, then
is a singular
-cube which is a coordinate system for
in a neighborhood
of
. Depending on the sign of
, the new
-cube is either
orientation preserving or reversing. In particular,
is the volume element of
if
is the volume element of
(which
is also orientable). Since the volume is just the integral of the volume
element and the integral is calculated via the
-cubes, it follows that
the volumes of
and
are equal.
This was already done in Problem 5-21.
show that
is a M@ouml;bius strip and find its area.
To see that it is a M@ouml;bius strip, note that in cylindrical coordinates,
the equations are:
. In particular,
for fixed
, we have
fixed and the path is a line segment
traversed from
to
. Calculating the length of the line,
one gets that is a line segment of length 2. Again, for fixed
, the
line segment in the
-plane has slope
. Note that this varies
from
down to
as
ranges from 0 to
, i.e. the line
segment starts vertically at
and reduces in slope until it becomes vertical
again at
. This corresponds to twisting the paper 180 degrees as it
goes around the ring, which is the M@ouml;bius strip.
To find the area, one can actually, just use the formulas for an orientable
surface, since one can just remove the line at
. In thatt
case one can verify, preferably with machine help, that
,
, and
. So the area is
. Numerical
evaluation of the integral yields the approximation
,
which is just slightly larger than
, the area of a circular ring of radius 2
and height 2.
Suppose
is the nowhere-zero
-form on
. If
is a singular
-cube, then for every
, we have
because the space
is of dimension 1. Choose
a
-cube so that the value is positive for some p. Then if it were negative
at another point
, then because this is a continuous function
from
into
, the intermediate value theorem would guarantee
a point
where the function were zero, which is absurd. So the value is
positive for all
.
For every
, choose a
-cube of this type with
in its image.
Define
. This is well defined. Indeed,
if we had two such
-cubes, say
and
, then the
(where
) are positive for
. But then the values are
of the same sign regardless of which non-zero k-form in
is used.
So, both maps define the same orientation on
.
This is an immediate consequence of Problem 5-23.
We will need
to be continuously differentiable, not just differentiable.
Following the hint, if
, then by the
mean value theorem,
for some
. Summing over
gives a Riemann sum
for
. Taking the limit as the mesh approaches 0,
shows that these approach the integral. Starting from any partition, and
taking successively finer partitions of the interval with the mesh approaching
zero, we get an increasing sequence of values with
limit the value of the integral; so the integral is the least upper bound of all
these lengths.
This is a straightforward calculation using the definition and Theorem 4-10.
For example,
. The
other two terms give similar results, and the sum is zero.
For
let
. Show that
restricted to the tangent space of
is
times the volume
element and that
. Conclude that
is not exact. Nevertheless, we denote
by
since, as we shall
see,
is the analogue of the 1-form
on
.
As in the proof of Theorem 5-6 (or Problem 5-25), the value of
can be evaluated by expanding
using cofactors
of the third row.
The second assertion follows from
and the fact
that the outward directed normal can be taken to be
be
an appropriate choice of orientation. One has
by Problem 5-26.
If
, then Stokes' Theorem would imply that
. Since the value is
, we conclude that
is not exact.
If
, then using part (b), one has
. By Problem 5-9, for any point
on the generalized cone, the line through
and the origin
lies on the surface and so its tangent line (the same line) is in
.
But then
is identically 0 for all points
. But then
.
We take the orientations induced from the usual orientation of
.
Following the hint, choose
small enough so that there is a three
dimensional manifold-with-boundary N (as in Figure 5-10) such that
is the union of
and
, and a part of a generalized
cone. (Actually,
will be a manifold-with-corners; see the remarks at the
end of the next section.)
Note that this is essentially the same situation as in Problem 5-22. Applying
Stokes' Theorem gives
because
is closed by part (a). By part (c), the integral over the
part of the boundary making up part of a generalized cone is zero. The
orientation of the part of the boundary on
is opposite to that
of the orientation of the same set as a part of
. So, we have
and the last integral is the
solid angle subtended by
by the interpretation of
of part (b).
}
This follows immediately from Problem 4-34 (b) and Problem 5-31 (a).
where
This follows by direct substitution using the expression for
in the preamble to Problem 31.
This follows from the formula in part (b) since the third column of the
determinant defining
is zero.
The curves of Figure 4-5(b) are given by
and
. You may easily convince yourself
that calculating
by the above integral is hopeless in this case.
The following problem shows how to find
without explicit
calculations.
If
is a compact two-dimensional manifold-with-boundary in
and
define
Let
be a point on the same side of
as the outward normal and
be a point on the opposite side. Show that by choosing
sufficiently close to
we can make
as close to
as desired.
Following the hint, suppose that
where
is a compact
manifold-with-boundary of dimension 3. Suppose
.
Removing a ball centered at
from the interior of
gives another
manifold-with-boundary
with boundary
, where the
orientation on
is opposite to that of the induced orientation.
So by Stokes' Theorem and Problem 5-31 (b),
. (Note the
discrepancy in sign between this and the hint.) On the other
hand, if
, then by Stokes' Theorem,
.
The rest of the proof will be valid only in the case where there is
a 3-dimensional compact oriented manifold-with-boundary
such that
where
is a two-dimensional manifold. Then
, and
and we can take
. So,
and
by the last paragraph. Subtracting gives
. The first term
can be made as small as we like by making
sufficiently small.
In the statement, what one means is that the component of
in the outward
normal direction is either in the same or opposite direction as the outward
normal.
Parameterize
with
where
is not in
. Let
be the values where
intersects
. To complete the proof, we will need to assume that
is finite. Let
. Choose
small enough
so that
for all
(where by
we mean
. One has
By part (a), if the tangent vector to
at
has a component in the
outward normal direction of
is positive, then
is
; if it is negative, then it is
. So, the last paragraph has
.
Note that this result differs from the problem statement by a sign.
where
.
The definition of
should be
.
The proofs are analogous; we will show the first result. Start with
where we have used Problem 3-32 to interchange the order of the limit and the integral.
On the other hand, we have by Stokes' Theorem that
Comparing the two expressions, we see that two of the terms in the first
expression match up with corresponding terms in the second expression. It
remains to check that the remaining terms are equal. But a straightforward
expansion gives:
as desired.
By part (c), one has
and so
. Similar results hold for
and
. One then substitutes into:
. After
collecting terms, we see that this is equal to the expression for
in Problem 5-32 (b). So,
.
By inspection in Figure 4-6, one has
and
in parts
(b) and (c) of the figure. So, these are also the values of
.