Since = p, it follows that
We can also write the integration limits for ^' immediately on observing the region % since for fixed <j>, p varies from p = 2 to p = 3 within the sector shown dashed in Fig. 6-10(a). An integration with respect to <j> from <j> = 0 to <f> = 2ir then gives the contribution from all sectors.
Geometrically pdpd<f> represents the area dA as shown in Fig. 6-10(a).
6.8. E x p r e s s \ d x d y d z in cylindrical coordinates. «
The transformation equations in cylindrical coordinates are x = p cos <f>, y = p sin <f>, z — z.
The Jacobian of the transformation is
Then by Problem 6.7 the triple integral becomes
where ^' is the region in the p, <p, z space corresponding to "•£ and where G(p, </>, «) = F(p cos 0, p sin <f>, z)
6.9. Find the volume of the region above the xy plane bounded by the paraboloid z = x
2+ y
2and the cylinder x
z+ y
2- a
2.
The volume is most easily found by using cylindrical coordinates. In these coordinates the equations for the paraboloid and cylinder are re- spectively z = p2 and p = a. ThenRequired volume
= 4 times volume shown in Pig. 6-12
The integration with respect to z (keeping p and <f> constant) from z = 0 to z = p2 corresponds to summing the cubical volumes (indicated by dV) in a vertical column extending from the xy plane to the paraboloid. The subsequent integration with respect to p (keeping <t> constant) from p = 0 to p = a corresponds to addition of volumes of all columns in the wedge shaped region. Finally, integration with respect to <f> corresponds to adding volumes of all such wedge shaped regions.
The integration can also be performed in other orders to yield the same result.
We can also set up the integral by determining the region ^' in p, $, z space into which ^ is mapped by the cylindrical coordinate transformation.
LINE INTEGRALS
5.10. Evaluate along (a) a straight line from (0,1) to (1,2), (6) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2), (c) the parabola x = t, y = t
2+ l.
(a) An equation for the line joining (0,1) and (1,2) in the xy plane is y = x + l. Then dy = dx and the line integral equals
Fig. 6-12
(b) Along the straight line from (0,1) to (1,1), y = 1, dy = 0 and the line integral equals
Along the straight line from (1,1) to (1,2), x = 1, dx = 0 and the line integral equals
Then the required value = -2/3 +10/3 = 8/3.
(c) Since í = 0 at (0,1) and f = 1 at (1, 2), the line integral equals
6.11. If A = (3a;
2-6ye)i + (2y + 3xz)j + (1 -4xyz
2)k, evaluate k-dr from (0,0,0) to (1,1,1) along the following paths C:
(a) x-t, y = t
2, z = t
3.
(b) the straight lines from (0,0,0) to (0,0,1), then to (0,1,1), and then to (1,1,1).
(c) the straight line joining (0,0,0) and (1,1,1).
(a) If x = t, y = t2, z = t3, points (0, 0,0) and (1,1,1) correspond to t = 0 and í = 1 respectively.
Then
Another method.
Along C, A = (3 «2 - 6i«)i + (2«2 + 3t*)j + (1 - 4i")k and r = x\ + yj + zk = tí. + t*j + t%, dr = (i + 2ij + 3f%) dt. Then
(b) Along the straight line from (0,0,0) to (0,0,1), x = O, y = 0, dx - 0, dy = 0 while z varies from 0 to 1. Then the integral over this part of the path is
Along the straight line from (0, 0,1) to (0,1,1), x = 0, z - 1, dx = 0, dz = 0 while y varies from 0 to 1. Then the integral over this part of the path is
Along the straight line from (0,1,1) to (1,1,1), y = 1, z = 1, dy = 0, dz = 0 while x varies from 0 to 1. Then the integral over this part of the path is
Adding,
(c) The straight line joining (0,0,0) and (1,1,1) is given in parametric form by * = t, y = t, z = t.
Then
6.12. Find the work done in moving a particle once around an ellipse C in the xy plane, if the ellipse has center at the origin with semi-major and semi-minor axes 4' and 3 respectively, as indicated in Fig. 6-13, and if the force field is given by
F = (3x-4y + 2z)i + (4x + 2y- 3z
2) j + (2xz - 4y* + «
3)k
In the plane 2 = 0, F = (3« - 4j/)i + (4* + 2y)j - 4î/2k and dr = dxi + dyj so that the work done is
Choose the parametric equations of the ellipse as x = 4 cos t, y = 3 sin í where t varies from 0 to 2ir (see Fig. 6-13). Then the
line integral equals Fig. 6-13
In traversing C we have chosen the counterclockwise direction indicated in Fig. 6-13. We call this the positive direction, or say that C has been traversed in the positive sense. If C were traversed in the clockwise (negative) direction the value of the integral would be — 96ir.
6.13. Evaluate y ds along the curve C given by from x = 3 to x = 24.
Since we have
GREEN'S THEOREM IN THE PLANE
6.14. Prove Green's theorem in the plane if C is a closed curve which has the property that any straight line parallel to the co- ordinate axes cuts C in at most two points.
Let the equations of the curves AEB and AFB (see adjoining Fig. 6-14) be y - Fj (a;) and y = Y2(x) respectively. If <% is the region bounded by C, we have
Similarly let the equations of curves EAF and EBF be x - X^y) and x = X2(y) respectively.
Then
Then <«>
Adding (1) and (2), dx dy.
6.15. Verify Green's theorem in the plane for
where C is the closed curve of the region bounded by y = x
2and y
2= x.
The plane curves y = x2 and y2 = * inter- sect at (0,0) and (1,1). The positive direction in traversing C is as shown in Pig. 6-15.
Along y = x2, the line integral equals Fig. 6-15
Along yz = x the line integral equals
Then the required line integral = 7/6 - 17/15 = 1/30.
Hence Green's theorem is verified.
6.16. Extend the proof of Green's theorem in the plane given in Problem 6.14 to the curves C for which lines parallel to the coordinate axes may cut C in more than two points.
Consider a closed curve C such as shown in the adjoining Fig. 6-16, in which lines parallel to the axes may meet C in more than two points. By constructing line ST the region is divided into two regions "^ and ^2 which are of the type con- sidered in Problem 6.14 and for which Green's
theorem applies, i.e., p¡g. 6-16
Then (1)
S'
Adding the left hand sides of (Í) and (2), we have, omitting the integrand P dx + Q dy in each case,
using the fact that
Adding the right hand sides of (1) and (2), omitting the integrand, where "^ consists of regions "^ and ^2-
A region "3J such as considered here for which any closed curve lying in "3^ can be continuously shrunk to a point without leaving <ï{, is called a simply-connected region. A region which is not simply-connected is called multiply-connected. We have shown here that Green's theorem in the plane applies to simply-connected regions bounded by closed curves. In Problem 6.19 the theorem is extended to multiply-connected regions.
For more complicated simply-connected regions it may be necessary to construct more lines, such as ST, to establish the theorem.
6.17. Show that the area bounded by a simple closed curve C is given by
In Green's theorem, put P = —y, Q = x. Then
where A is the required area. Thus ¡
6.18. Find the area of the ellipse x — a cos 0, y = b sin 0.
Area [(a cos e)(b cos e) de — (b sin »)(—a sin e) de]
6.19. Show that Green's theorem in the plane is also valid for a multiply-connected region "3Ç.
such as shown in Fig. 6-17.
The shaded region *R_, shown in the figure, is multiply-connected since not every closed curve lying in ^ can be shrunk to a point without leaving <£, as is observed by considering a curve surrounding DEFGD for example. The boundary of ^, which consists of the exterior boundary AHJKLA and the interior boundary DEFGD, is to be traversed in the positive direction, so that a person traveling in this direction always has the region on his left. It is seen that the positive directions are those indicated
in the adjoining figure. Fig. 6-17
In order to establish the theorem, construct a line, such as AD, called a cross-cut, connecting the exterior and interior boundaries. The region bounded by ADEFGDALKJHA is simply-con- nected, and so Green's theorem is valid. Then
( 1 )
d x d y a n d t h e t h e o r e m i s p r o v e d .
Then
figure.
But the integral on the left, leaving out the integrand, is equal to
since . Thus if Ct is the curve ALKJHA, C2 is the curve DEFGD and C is the boundary of "3? consisting of C1 and C2 (traversed in the positive directions), then
and so -