Let S be an open, two-sided surface bounded by a closed non-intersecting curve C (simple closed curve). Consider a directed line normal to S as positive if it is on one side of S, and negative if it is on the other side of S. The choice of which side is positive is arbitrary but should be decided upon in advance. Call the direction or sense of C posi- tive if an observer, walking on the boundary of <S with his head pointing in the direction of the positive normal, has the surface on his left. Then if Ai,A
2, A
3are single-valued, continuous, and have continuous first partial derivatives in a region of space including S, we have
(38)
In vector form with A = Aii + A
2j + A
3k and n = cos ai + cos/îj + cos-yk, this is simply expressed as
(39)
In words this theorem, called Stokes' theorem, states that the line integral of the
tangential component of a vector A taken around a simple closed curve C is equal to the
surface integral of the normal component of the curl of A taken over any surface S
having C as a boundary. Note that if, as a special case V x A = 0 in (39), we obtain
the result (28).
Solved Problems
DOUBLE INTEGRALS
6.1. (a) Sketch the region 9?. in the xy plane bounded by y = x
2, x = 2, y = 1.
(&) Give a physical interpretation to (c) Evaluate the double integral in (&).
(a) The required region ^ is shown shaded in Fig. 6-4 below.
(b) Since x2 + y2 is the square of the distance from any point (x, y) to (0,0), we can consider the double integral as representing the polar moment of inertia (i.e. moment of inertia with respect to the origin) of the region ^. (assuming unit density).
We can also consider the double integral as representing the mass of the region "3{ assuming a density varying as x2 + y2.
Fig. 6-4 Fig. 6-5
(c) Method 1. The double integral can be expressed as the iterated integral
The integration with respect to y (keeping x constant) from y — 1 to y — x2 corresponds formally to summing in a vertical column (see Fig. 6-4). The subsequent integration with respect to x from x = 1 to x = 2 corresponds to addition of contributions from all such vertical columns between x — 1 and x = 2.
Method 2. The double integral can also be expressed as the iterated integral
In this case the vertical column of region ^ in Fig. 6-4 above is replaced by a horizontal column as in Fig. 6-5 above. Then the integration with respect to x (keeping y constant) from x = i/y to x = 2 corresponds to summing in this horizontal column. Subsequent integration with respect to y from y - Í to y = 4 corresponds to addition of contributions for all such horizontal columns between y = 1 and y = 4.
6.2. Find the volume of the region common to the intersecting cylinders x
z+ y
2= a
2and x
2+ z
2= a
2.
Required volume = 8 times volume of region shown in Fig. 6-6
As an aid in setting up this integral note that z dy áx corresponds to the volume of a column such as shown darkly shaded in the figure. Keeping x constant and integrating with respect to y from y = 0 to y = ^az — x^ corresponds to adding the volumes of all such columns in a slab parallel to the yz plane, thus giving the volume of this slab. Finally, integrating with respect to
* from x = 0 to x = a, corresponds to adding the volumes of all such slabs in the region, thus giving the required volume.
Fig. 6-6 Fig. 6_7
TRIPLE INTEGRALS
6.3. (a) Sketch the 3 dimensional region ^ bounded by x + y + z = a(a,>0), x = 0 y = Q
2 = 0. '
(&) Give a physical interpretation to
(c) Evaluate the triple integral in (6).
(a) The required region ^ is shown in Fig. 6-7.
(6) Since x2 + y* + 32 is the square of the distance from any point (x, y, z) to (0, 0,0), we can consider the triple integral as representing the polar moment of inertia (i.e. moment of inertia with respect to the origin) of the region ^ (assuming unit density).
We can also consider the triple integral as representing the mass of the region if the density varies as a2 + j/2 + z2.
(c) The triple integral can be expressed as the iterated integral
Aims me centróla ñas coordinates (3/4,3,8/5).
Note that the value for y could have been predicted because of symmetry.
Total mass
Total moment about xy plane Total moment about xz plane
Total mass
Total moment about yz plane Total mass
by part (a), since a is constant. Then Fig. 6-8
(6) Total mass (a) Required volume
The region ^ is shown in Fig. 6-8.
6.4. Find the (a) volume and (b) centroid of the region <5? bounded bv the narahnlip cylinder z = 4-x
¿and the planes x = Q, y = 0, y = Q,z^Q assuming the density to be a constant a.
The integration with respect to z (keeping * and y constant) from z = 0 to z = a-x-y corresponds to summing the polar moments of inertia (or masses) corresponding to each cube in a vertical column. The subsequent integration with respect to y from y = 0 to y = a — x (keeping * constant) corresponds to addition of contributions from all vertical columns contained in a slab parallel to the yz plane. Finally, integration with respect to x from * = 0 to x = a adds up contributions from all slabs parallel to the yz plane.
Although the above integration has been accomplished in the order z, y, x, any other order is clearly possible and the final answer should be the same.
TRANSFORMATION OF DOUBLE INTEGRALS 6.5. Justify equation (21), page 151, for chang-
ing variables in a double integral.
In rectangular coordinates, the double integral oí' F(x,y) over the region ^ (shaded in Fig. 6-9) is We can also evaluate this double integral by considering a grid formed by a family of u and v curvilinear coordinate curves constructed
on the region ••£ as shown in the figure.
Let P be any point with coordinates (x, y) or (u,v), where x = f(u,v) and y = g(u,v). Then the vector r from O to P is given by r = xi + yj = f(u, v)i + g(u, v)j. The tangent vectors to the coordi- nate curves u — ct and v = cz, where Cj and C2 are constants, are dr/dv and dr/du respectively. Then the area of the region A^ of Fig. 6-9 is given ap-
proximately by AwAt>. Fig. 6-9 But
so that
The double integral is the limit of the sum
taken over the entire region "3Ç. An investigation reveals that this limit is
where <^' is the region in the uv plane into which the region "^ is mapped under the transformation•\.
x = f(u, v), y = g(u, v).
6.6. Evaluate where 'K. is the region in the xy plane bounded by
£2 + 3,2 = 4 and x
2+ y
2= 9.
The presence of xz + y2 suggests the use of polar coordinates (p, 0), where » = p cos <f>, y = p sin <f>.
Under this transformation the region "^ [Fig. 6-10(a)] is mapped into the region <3(' [Fig. 6-10(6)].
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).