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, 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).
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.
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 .