Part 21 Triple Integrals Divergence Theorem of Gauss-Advanced Engineering Mathematics-Solution Manual, Exercises of Engineering Mathematics

Download Part 21 Triple Integrals Divergence Theorem of Gauss-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! 14. x  0 because of symmetry. M  1_ 2a 2. For y, using polar coordinates defined by x  r cos , y  r sin , we obtain y 
 0
a 0 (r sin ) r dr d 
 0 sin d 

2   0.4244a. For a  1 we get 4/(3), as in Example 2. It is obvious that the present semidisk and the quarterdisk in the example have the same y. 16. x  b/2 for reasons of symmetry. Since the given R and its left half (the triangle with vertices (0, 0), (b/2, 0), (b/2, h)) have the same y, we can consider that half, for which M  1_4bh. We obtain y 
b/2 0
2hx/b 0 y dy dx 
b/2 0 ( )2 dx 
( )2
( )3  . This is the same value as in Prob. 15. This equality of the y-values seems obvious. 18. We obtain Ix 
b/2 0
2hx/b 0 y2 dy dx 
b b/2
2h
2hx/b 0 y2 dy dx 
b/2 0 ( )3 dx 
b b/2 [2h(1  )]3 dx  bh3  bh3  bh3. Each of those two halves of R contributes half to the moment Ix about the x-axis. Hence we could have simplified our calculation and saved half the work. Of course, this does not hold for Iy. We obtain Iy 
b/2 0
2hx/b 0 x2 dy dx 
b b/2
2h
2hx/b 0 x2 dy dx 
b/2 0 x2 ( ) dx 
b b/2 x2 (2h  ) dx  b3h  b3h  b3h. 20. We denote the right half of R by R1  R2, where R1 is the rectangular part and R2 the triangular. The moment of inertia Ix1 of R1 with respect to the x-axis is Ix1 
b/2 0
h 0 y2 dy dx 
b/2 0 dx  bh3. 1  6 h3  3 7  48 11  96 1  32 2hx  b 2hx  b 1  12 1  24 1  24 x  b 1  3 2hx  b 1  3 h  3 b  2 1  3 2h  b 1  2 4  bh 2hx  b 1  2 4  bh 4  bh 4a  3 a3  3 2  a2 a3  3 1  M 1  M Instructor’s Manual 195 im10.qxd 9/21/05 12:49 PM Page 195 docsity.com Similarly for the triangle R2 we obtain Ix2 
a/2 b/2
h(2x
a) /(b
a) 0 y2 dy dx 
a/2 b/2 dx  h3(a  b). Together, Ix  (3b  a) and Ix  h 3(3b  a). Iy is the same as in Prob. 19; that is, Iy  . This can be derived as follows, where we integrate first over x and then over y, which is simpler than integrating in the opposite order, where we would have to add the two contributions, one over the square and the other over the triangle, which would be somewhat cumbersome. Solving the equation for the right boundary y  (a  2x) for x, we have x  (ah  (a  b)y) and thus Iy 
h 0
(ah
(a
b)y) /2h 0 x2 dx dy 
h 0 (ah  (a  b)y)3 dy  (a3  a2b  ab2  b3)  . Now we multiply by 2, because we considered only the right half of the profile. SECTION 10.4. Green’s Theorem in the Plane, page 439 Purpose. To state, prove, and apply Green’s theorem in the plane, relating line and double integrals. Comment on the Role of Green’s Theorem in the Plane This theorem is a special case of each of the two “big” integral theorems in this chapter, Gauss’s and Stokes’s theorems (Secs. 10.7, 10.9), but we need it as the essential tool in the proof of Stokes’s theorem. The present theorem must not be confused with Green’s first and second theorems in Sec. 10.8. h(a4  b4)  96(a  b) h  96 1  24h3 1  2 1  2h h  a  b h(a4  b4)  48(a  b) 1  12 h3  24 1  2 1  24 h3(2x  a)3  (b  a)3 1  3 196 Instructor’s Manual im10.qxd 9/21/05 12:49 PM Page 196 docsity.com SOLUTIONS TO PROBLEM SET 10.5, page 448 2. Circles, straight lines through the origin. A normal vector is N  k k  [0, 0, u]  uk. At the origin this normal vector is the zero vector, so that (4) is violated at (0, 0). This can also be seen from the fact that all the lines v  const pass through the origin, and the curves u  const (the circles) shrink to a point at the origin. This is a consequence of the choice of the representation, not of the geometric shape of the present surface (in contrast with the cone, where the apex has a similar property, but for geometric reasons). 4. x2  y2  z, circles, parabolas with the z-axis as axis; a normal vector is [2u2 cos v, 2u2 sin v, u]. 6. 1_4x 2  y2  z, hyperbolas, parabolas. A normal vector is [2u2 cosh v, 8u2 sinh v, 4u]. 8. z  arctan (y/x), helices (hence the name!), horizontal straight lines. This surface is similar to a spiral staircase, without steps (as in the Guggenheim Museum in New York). A normal vector is [sin v, cos v, u]. 10. x2/a2  y2/b2  z2/c2  1. Both families of parameter curves consist of ellipses. A normal vector is [bc cos2 v cos u, ac cos2 v sin u, ab sin v cos v]. 12. z  5_3x  1_ 3y  10; hence [u, v, 5_3u  1_3v  10], or if we multiply by 3, r(u, v)  [3u, 3v, 5u  v  30]. A normal vector is N  k k  [15, 3, 9]. More simply, [5, 1, 3] by applying grad to the given representation. The two vectors are proportional, as expected. 14. In (3) the center is (0, 0, 0). Here it is (1, 2, 0). Hence we obtain r(u, v)  [1  5 cos v cos u, 2  5 cos v sin u, 5 sin v]. From this and (4) we obtain the normal vector N  [25 cos2 v cos u, 25 cos2 v sin u, 25 cos v sin v]. 16. A parametric representation is r(u, v)  [u cos v, 2u sin v, 4u2]. A normal vector is N  [16u2 cos v, 8u2 sin v, 2u]. k 5 1 j 0 3 i 3 0 k 0 0 j sin v u cos v i cos v u sin v Instructor’s Manual 199 im10.qxd 9/21/05 12:49 PM Page 199 docsity.com Another one is grad (z  4x2  y2)  [8x, 2y, 1]  [8u cos v, 4u sin v, 1]. Multiplication by 2u gives the previous normal vector. 18. A representation is r(u, v)  [2 cosh u, 3 sinh u, v]. A normal vector is N  [3 cosh u, 2 sinh u, 0]. Note that N is parallel to the xy-plane. Another normal vector is grad (9x2  4y2)  [18x, 8y, 0]  [36 cosh u, 24 sinh u, 0]. Division by 12 gives the previous normal vector. 20. Set x  u and y  v. 22. N(0, 0)  0 in Prob. 2 (polar coordinates); see the answer to Prob. 2. In Probs. 4 and 7 (paraboloids) the situation is similar to that of the polar coordinates. In Prob. 6 the origin is a saddle point. In each of these cases one can find a representation for which N(0, 0) 0; see Prob. 23 for the paraboloid. In Prob. 5 the reason is the form of the surface (the apex of the cone, where no tangent plane and hence no normal exists). 24. Project. (a) ru(P) and rv(P) span T(P). r* varies over T(P). The vanishing of the scalar triple product implies that r*  r(P) lies in the tangent plane T(P). (b) Geometrically, the vanishing of the dot product means that r*  r(P) must be perpendicular to g, which is a normal vector of S at P. (c) Geometrically, ƒx(P) and ƒy(P) span T(P), so that for any choice of x*, y* the point (x*, y*, z*) lies in T(P). Also, x*  x, y*  y gives z*  z, so that T(P) passes through P, as it should. SECTION 10.6. Surface Integrals, page 449 Purpose. We define and discuss surface integrals with and without taking into account surface orientations. Main Content Surface integrals (3)  (4)  (5) Change of orientation (Theorem 1) Integrals (6) without regard to orientation; also (11) Comments on Content The right side of (3) shows that we need only N but not the corresponding unit vector n. An orientation results automatically from the choice of a surface representation, which determines ru and rv and thus N. The existence of nonorientable surfaces is interesting but is not needed in our further work. SOLUTIONS TO PROBLEM SET 10.6, page 456 2. r  [u, v, 4  u  v], 0  u  4  v, 0  v  4, F  [u2, v2, (4  u  v)2], N  [1, 1, 1]. From this we obtain
4 0
4
v 0 F • N du dv 
4 0 [1_3u3  v2u  1_3(4  u  v)3]j 0 4
v dv  64. 200 Instructor’s Manual im10.qxd 9/21/05 12:49 PM Page 200 docsity.com 4. Quarter of a circular cylinder of radius 3 and height 2 in the first octant with the z-axis as axis. A parametric representation is r  [3 cos u, 3 sin u, v], 0  u  1_2, 0  v  2. From this we obtain N  [3 cos u, 3 sin u, 0] F • N  3e3 sin u cos u  3ev sin u. Integration over u from 0 to 1_2 gives e 3  1  3ev. Integration of this over v from 0 to 2 gives the answer 2(e3  1)  3(e2  1)  2e3  3e2  1  19.0039. 6. r  [u, cos v, sin v], 0  u  20, 0  v  , N  [0, cos v, sin v]; hence F • N  cos4 v sin v. Integration over v from 0 to  gives 2/5. Integration of this over u from 0 to 20 gives the answer 8. 8. r  [u, cos v, 3 sin v], N  [0, 3 cos v, sin v]; hence F  [tan(u cos v), u2 cos v, 3 sin v]. This gives F • n  3u2 cos2 v  3 sin2 v. Integration over u from 1 to 4 gives 63 cos2 v  9 sin2 v. Integration of this over v from 0 to 2 gives the answer 63  9  54. 10. Portion of a circular cone with the z-axis as axis. A parametric representation is r  [u cos v, u sin v, 4u] (0  u  2, 0  v  ). From this, N  [4u cos v, 4u sin v, u], F  [u2 sin2 v, u2 cos2 v, 256u4]. The integrand is F • n  4u3 sin2 v cos v  4u3 cos2 v sin v  256u5. Integration over u from 0 to 2 gives 16 sin2 v cos v  16 cos2 v sin v  . Integration of this over v from 0 to  gives [ sin3 v  cos3 v  v] 0      8568. 12. r  [u, v, u  v2], N  [1, 2v, 1], F  [cosh v, 0, sinh u]. Hence the integrand is F • N  cosh v  sinh u. 8192  3 32  3 8192  3 16  3 16  3 8192  3 Instructor’s Manual 201 im10.qxd 9/21/05 12:49 PM Page 201 docsity.com