quiz2B-2011 - The University of Sydney School of...

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Unformatted text preview: The University of Sydney School of Mathematics and Statistics Solutions to Vector Calculus - Quiz B MATH2061/2067: 1. Find a potential function of the gradient field F = (y 2 - (sin x)ez ) i + 2xy j + (cos x)ez k. (a) xy 2 + (cos x)ez (c) 2xy - (sin x)ez Solution: (a) 2. Which of the following integrals gives the length of the curve r t = cos t i + sin t j + t2 k, 1 2 Semester 1, 2011 (b) 2xy + (sin x)ez - (cos x)ez (d) xy 2 + (sin x)ez . t : 0 2 ? (a) 0 2 1 + 4t2 dt 1 + 4t2 dt (b) 0 1 1 + 4t2 dt 1 + 4t2 dt. (c) 0 (d) 0 Solution: (c) 3. Let F = 2x i + y j. Find the flux C F n ds, where C is the unit circle, centre (0, 0), taken once anti-clockwise, and n the unit outward normal. (a) 0 Solution: (c) 4. Let F = (2ax + y) i + (2y + z) j + (x - 6az) k, where a is a real constant. Find any values of a for which F = 0. 1 1 (a) a = (b) a = 4 2 (c) F = 0 for all values of a. (d) F is not zero for any a. Solution: (b) (b) /2 (c) 3 (d) 2. Copyright c 2011 The University of Sydney 1 =/2 =/2 5. Evaluate =0 =0 cos3 sin dd. (b) 8 (c) - 6 (d) - . 8 (a) 6 Solution: (b) 6. Which of the following integrals gives the volume of the solid enclosed between the paraboloid z = 1 - x2 - y 2 , z 0, and the xy-plane ? =2 r=1 =2 r=1 (a) =0 =2 r=0 r=1 r=0 r dr d (1 - r 2 )r dr d 2 (b) =0 =2 r=0 r=1 r=0 (1 - r 2 )1/2 r dr d (1 - r 2 )1/2 dr d. (c) =0 (d) =0 Solution: (c) 7. The double integral of the function = x2 + 2y over the region of the xy plane in the first quadrant, bounded by the curves y = 1 - x2 , y = 0 and x = 0, is given by: 1 1-x2 (a) 0 1 0 1 x2 + 2y dx dy x2 + 2y dy dx 0 1 1-x2 1-y (b) (c) 0 1 1 0 x2 + 2y dy dx 1-y (d) 0 x2 + 2y dx dy . Solution: (d) 8. Evaluate gram. y 1 (x + 2y) dA where R is the shaded region shown in the diaR 1 x (a) 1/2 (b) 4/3 (c) 3/2 (d) 2/3. Solution: (a) 2 9. Find a unit vector normal to the surface e2x y+xz 2 = 1 at the point (0, 1, 1). i + 3j 3i + j 3i + j + k i + 3j + k (a) (b) (c) (d) . 10 10 11 11 Solution: (b) 10. Calculate (x2 y i + 2yz j + xz 2 k). (a) -2y i - z 2 j - x2 k (c) -2y i + z 2 j - x2 k (d) -2y i - z 2 j + x2 k. (b) 2y i - z 2 j - x2 k Solution: (a) 11. If F is a conservative vector field, one of the following statements is not necessarily true. Which one is not true? (a) C F dr = 0 for any closed curve C (b) F = 0 (d) F = 0. (c) F = for some scalar function Solution: (b) 12. The equation of the tangent plane to the surface z = x2 y + y 2 at the point (1, 1, 2) is: (a) 2x + 3y - z = 2 (c) 2x + 3y - z = 3 (d) 2x + 3y - z = 4. (b) 2x - 3y - z = 3 Solution: (c) 13. Suppose F is a vector field, and F = (xy 2 + zy + xz 2 ). Let C be the straight line segment from (0, 0, 1) to (1, 1, 2). Evaluate (a) 0 Solution: (b) 14. Let C be the parabola y = x2 from the point (0, 0) to the point (1, 1). Evaluate C C F d r. (b) 7 (c) 3 (d) 5. F d r, where F = x i - y 2 j. (b) 1/3 (c) 1/6 (d) 2/3. (a) 1/12 Solution: (c) 3 15. Let F be a vector field and be a scalar function (both differentiable as required). Then (F) (a) is zero if F is conservative (c) is a scalar quantity Solution: (c) (b) is a vector quantity (d) is not a meaningful expression. THIS IS THE END OF THE QUESTIONS. 4 Formula Sheet Most of the formulas and theorems provided are stated without the conditions under which they apply. The notation used is the same as that used in lectures. Line Integrals b (x, y, z) ds = C a (x(t), y(t), z(t)) dx dt 2 + dy dt 2 + dz dt 2 dt. C F dr = F1 dx + F2 dy + F3 dz = Work done by F along C. C Grad grad = = i+ j+ k. x y z is normal to the tangent plane of the surface (x, y, z) = k, (k a constant). If F is continuous and equal to for some , then F is a conservative field, is a potential function of F, and C F dr is path-independent. Curl i curl F = F = x F1 j y F2 k z F3 If the domain of F is simply connected and F = 0 then F is conservative. Double integrals over a region R in R2 . Area of R = R dA. f (x, y) dA. R Volume of the solid under the surface z = f (x, y), over R = In polar coordinates: R f (x, y) dA = R f (r cos , r sin )r d dr. Divergence div F = F = F1 F2 F3 + + . x y z 5 Green's theorem F dr = F1 dx + F2 dy = C R C F2 F1 - x y dx dy . In vector form: C F dr = C R ( F) k dA. R In divergence form: Surface integrals F n ds = F dx dy = Flux of F across C. (x, y, z) dS = S R 2 2 (x, y, f (x, y)) 1 + fx + fy dx dy In cylindrical coordinates: In spherical coordinates: f (x, y, z) dS = Flux across S = S (x, y, z) dS = (a cos , a sin , t) a d dt. f (a cos sin , a sin sin , a cos ) a2 sin d d . F n dS. Triple integrals In cylindrical coordinates: (x, y, z) dx dy dz = In spherical coordinates: f (x, y, z) dx dy dz = Divergence theorem F n dS = F dV f r cos sin , r sin sin , r cos r 2 sin dr d d r cos , r sin , t r dr d dt S V Stokes' theorem C F dr = S ( F) n dS 6 Table of Standard Integrals 1. xn dx = xn+1 +C n+1 (n = -1) 6. sec2 x dx = tan x + C 2. dx = ln |x| + C x 7. sinh x dx = cosh x + C ex dx = ex + C 8. cosh x dx = sinh x + C dx = sin-1 2 - x2 a 3. 4. sin x dx = - cos x + C 9. cos x dx = sin x + C dx = + x2 x a +C 5. 10. a2 1 a tan-1 x a +C 11. dx = sinh-1 2 + a2 x dx = cosh-1 2 - a2 x x a + C = ln x + x2 + a2 + C 12. x a +C (x > a) (x > a or x < -a) = ln x + x2 - a2 + C THIS IS THE END OF THE QUIZ PAPER. 7 ...
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