# vcprac06s - The University of Sydney School of Mathematics...

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Unformatted text preview: The University of Sydney School of Mathematics and Statistics Solutions to Practice Session 6 MATH2061: Vector Calculus 1. Evaluate S Summer School 2012 F n dS, where F = 2x i+3xy j-yz 2 k and S is the surface bounded by the planes x = 0, x + y = 2, y = 0, z = 0 and z = 2. ( n is the unit outward normal to the surface S.) z 2 Solution: Let V be the solid enclosed by S. Then V is described by: 0y 2-x 0z2 0x2 x 2 2 y Now, div F = F = 2 + 3x - 2yz, and so by the divergence theorem, S F n dS = = V F dV (2 + 3x - 2yz) dV 2 2-x 0 2 V 2 = 0 2 0 (2 + 3x - 2yz) dy dx dz = 0 2 0 ((2 + 3x)(2 - x) - (2 - x)2 z) dx dz 2 2 3 = 0 2 (2 - x)3 4x + 2x - x + z 3 8- 8z 3 dz = 32 . 3 dz 0 = 0 2. Find the flux of F = xz 2 i + (x2 y - z 3 ) j + (2xy + y 2 z) k outwards across the entire surface of the hemispherical region bounded by z = (a2 - x2 - y 2 )1/2 and z = 0. Solution: Let S be the entire surface of the given region, and let V be the solid enclosed by S. Then z S F n dS = = V F dV (z 2 + x2 + y 2 ) dV . x a y V Copyright c 2012 The University of Sydney 1 Using the spherical coordinates x = r cos sin , y = r sin sin , z = r cos , V is described by: 0 r a, 0 2, 0 /2 . Now, x2 + y 2 + z 2 = r2 , and dV = r2 sin dr d d. So, the flux = S F n dS 2 0 0 a /2 = 0 r4 sin dr d d 2a5 . 5 = 3. Evaluate S ( F) n dS , where S is the surface of the hemisphere x2 + y 2 + z 2 = a2 for z 0, F = y i + zx j + y k and n is the unit normal with positive k-component. (Note that there are various ways you could do this. For example, you could calculate the integral directly, or use the divergence theorem, or use Stokes' theorem. You should use whichever method you think will be the easiest.) Solution: The rim of S is the circle x2 + y 2 = a2 in the plane z = 0. This circle is also the rim of the disc x2 + y 2 a2 in the plane z = 0. Call the circle C, and the disc B. Then by Stokes' theorem, x ( F) n dS = F dr = y dx + zx dy + y dz, C z n y a S C where C is taken in the anti-clockwise direction. Since the upwards pointing normal to B is k, we also have (by two applications of Stokes' theorem) ( F) n dS = F dr = ( F) k dS. S C B Note that this result also follows from Gauss' theorem: On both S and B choose the normal outwards to the solid bounded by the closed surface S B. Then, by the divergence theorem, ( F) n dS = 2 ( F) dV = 0 . SB V But SB ( F) n dS = ( F) n dS = i y S ( F) n dS + ( F) ( k) dS. j zx k y B ( F) (- k) dS, so that S B Now F = /x /y /z = (1 - x) i + (z - 1) k, so that ( F) k = z - 1 = -1 (since z = 0 on B). Therefore S ( F) n dS = B ( F) n dS = (-1)dS B = -(area of the disc B) = -a2 . Alternately, evaluating the line integral above, we have C given by x = a cos , y = a sin , z = 0, : 0 2, and S ( F) n dS = = y dx + zx dy + y dz C 2 y dx = C 2 0 (a sin )(-a sin ) d = -a2 4. Evaluate C 0 sin2 d = -a2 . x2 y 3 dx + dy + z dz , where C is x2 + y 2 = a2 , z = 0, taken once, in an anti-clockwise direction when viewed from above. Solution: The evaluation of this integral is a little easier using Stokes' theorem, rather than integrating directly. Note that C is in the plane z = 0, so that x2 y 3 dx + dy + z dz = C C x2 y 3 dx + dy , and in this case Stokes' theorem is the same as Green's theorem. Let F = x2 y 3 i + j + z k . Then F = -3x2 y 2 k . Now choose some surface which has C as its rim. The simplest such surface is the disc x 2 + y 2 a2 , z = 0 . Call this disc S, and note that, since C is taken anti-clockwise, we choose the normal pointing up from S to use in Stokes' theorem. That is, we take n = k, 3 x z k C y and so ( F) n = -3x2 y 2 . We therefore have x2 y 3 dx + dy + z dz = C C F dr ( F) n dS -3x2 y 2 dS a 0 = S = S 2 = 0 -3r2 cos2 r2 sin2 rdr d 2 1 = - a6 2 1 = - a6 2 1 = - a6 8 cos2 sin2 d 0 2 0 2 0 sin2 2 d 4 6 1 - cos 4 d = -a 8 . 2 5. By using a suitable integration theorem, or otherwise, evaluate the line integral y dx + z dy + x dz C where C is the intersection of the sphere x2 +y 2 +z 2 = a2 and the plane x+y+z = 0. Assume that C is oriented anticlockwise, when viewed from a point on the z-axis with z > 0. Solution: The curve C is as shown: S C n The intersection C is a circle of radius a lying in the plane x + y + z = 0. Let F = y i+z j+x k and let S be the region bounded by C in the plane x+y +z = 0. Using Stoke's theorem, y dx + z dy + x dz = C C F dr ( F) n dS = S 4 where n is the unit normal to S with positive k-component. Now i F= x y j y z k = - i - j - k. z x On S1 , the unit normal is the normal to the plane x + y + z = 0: n= so that i+ j+ k 3 3 ( F) n = - . 3 y dx + z dy + x dz = C Hence 3a2 3 - dS = - . 3 3 S 5 ...
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