e x2 dxdy, e x2 da, e x2 x 3 dx = e

Transcription

1 STS26-4 Calculus II: The fourth exam Dec 15, 214 Please show all your work! Answers without supporting work will be not given credit. Write answers in spaces provided. You have 1 hour and 2minutes to complete this exam. Name: Student ID: 1. Evaluate the following integral Ans. We consider the followings: 1 3 3y 1 3 3y e x2 dxdy, e x2 dxdy D e x2 da, where D {(x, y) R 2 : y 1, 3y x 3}. So we can give an alternative description of D, Thus the above integral is {(x, y) R 2 : x 3, y 1 3 x} y e x2 dxdy 3 x/3 3 e e x2 x 3 dx e x2 dydx +1 : if you write the equation well +1 : if you find the critical points of this equation well +1 : if you give the three positive numbers

2 STS26-4 The fourth exam, Page 2 of 1 Dec 15, Find the volume of the solid lying under the elliptic paraboloid x 2 /4 + y 2 /9 + z 1 and above the rectangle R [ 1, 1] [ 2, 2]. Ans. Observe that the solid lying under z 1 x 2 /4 y 2 /9 and above R [ 1, 1] [ 2, 2]. So it is obvious that z 1 x 2 /4 y 2 /9 on (x, y) R. The volume of it is ) 1 (1 x2 4 y2 dydx 9 1 [4 x ] dx : if you write the equation well +2 : if you find the answer well

3 STS26-4 The fourth exam, Page 3 of 1 Dec 15, Evaluate the double integral. arctan(y/x)da, where R {(x, y) : 1 x 2 + y 2 4, y x} R Ans. In polar coordinates R is given by 1 r 2 and θ π 4. Moreover, arctan(y/x) arctan ( r cos θ r sin θ ) θ. So we have R arctan(y/x)da π/4 2 π/ π2. 1 θ rdrdθ θ 1 2 ( )dθ +1 : if you write R by the polar coordinate +1 : if you give arctan(y/x) +1 : if you find the answer well

4 STS26-4 The fourth exam, Page 4 of 1 Dec 15, xydV, where R lies under the plane z 1 + x + y and above the region in the E xy-plane bounded by the curves y x, y, and x 1. Ans. The solid E is shown in the following figure: F igure Then one of the projection of E onto xy-plane is D {(x, y) R 2 : x 1, y x}. So we have E {(x, y, z) R 3 : x 1, y x, z 1 + x + y}. The triple integral is E 6xydV 1 x 1+x+y 1 x 1 6xydzdydx 6xy(1 + x + y)dydx (3x 2 + ex 3 + 2x 5/2 dx [ x x4 + 4 ] 1 7 x7/ : if you express the equation well +2 : if you find the answer well

5 STS26-4 The fourth exam, Page 5 of 1 Dec 15, Evaluate B (x2 + y 2 + z 2 ) 2 dv, where B is the ball with center the origin and radius 5. Ans. We use spherical coordinates: B {(ρ, θ, φ) : ρ 5, θ 2π, φ π}. So the triple integral is (x 2 + y 2 + z 2 ) 2 dv B 5 2π π 5 2π 5 4πρ 6 dρ ρ 4 ρ 2 sin φdφdθdρ 2ρ 6 dθdρ 4 7 π : if you express the equation well +2 : if you find the answer well

6 STS26-4 The fourth exam, Page 6 of 1 Dec 15, (i) Find the image of the region R under the given transformation. R is the parallelogram with vertices ( 1, 3), (1, 3), (3, 1) and (1, 5); x 1(u + v), y 1 (v 3u) 4 4 (ii) Use the given transformation in (i) to evaluate the following integral. (4x + 8y)dA R Ans. (i) The transformation maps the boundary of R into the boundary of the image. Let R 1, R 2, R 3 and R 4 be the side of R as follows: F igure On R 1 : y 3x + 8, where 1 x 3, 1 y 5 Since v 3u 3 u+v + 8 and 1 u+v 3, so we have v 8, and 4 u On R 2 : y x + 4, where 1 x 1 Since v 3u u+v + 4 and 1 u+v 1, so we have u 4, and v On R 3 : y 3x, where 1 x 1 Since v 3u 3 u+v and 1 u+v 1, so we have v, and 4 u On R 4 : y x 4, where 1 x 3 Since v 3u u+v 4 and 1 u+v 4, so we have u 4, and v (ii) Use (i), we have 4 8 ( (4x + 8y)dA 4 u + v + 8 v 3u ) x y 4 4 u v. Note that So the above double integral is R 4 x y x u v u y u ( 5u + 3v) 1 4 dvdu 1 4 x v y v [ 5uv v2 ] 8 du )du 4 4( 4u [ 2u u ] 4 +2 : if you give the answer of (i) +2 : if you give the answer of (ii) 7. Evaluate the line integral xy 4 ds, where C is the right half of the circle x 2 + y 2 16 C 4

7 STS26-4 The fourth exam, Page 7 of 1 Dec 15, 214 Sol. Let x 4 cos t and y 4 sin t, where π t π/2. So ds 16 sin 2 t + 16 cos 2 t 4. Note that (dx ) 2 ( ) 2 ds dy dt + 4 dt dt 2 ( sin t) (cos t) 2 4. So this integral is C xy 4 ds π/2 π/2 4 cos t (4 sin t) 4 4dt +1 : if you give the equation by using the polar coordinate +2 : if you give the answer [ ] π/2 4 6 sin5 t π/2 5.

8 STS26-4 The fourth exam, Page 8 of 1 Dec 15, Evaluate the line integral C F dr. F(x, y, z) sin xi + cos yj + xzk, r(t) t 3 i t 2 j + tk, t 1 Ans. F C 1 1 F(r(t)) r (t)dt [ t sin(t 3 )3t 2 cos( t 2 )2t + t 4 dt cos(t 3 ) + sin( t 2 5 ) + t cos(1) + sin( 1) cos(1) sin(1). 5 ] 1 +1 : if you give the equation by using the parametric equation +1 : if you give the answer

9 STS26-4 The fourth exam, Page 9 of 1 Dec 15, Determine whether or not F is a conservative vector field. If it is, find a function f such that F f. F(x, y) (ye x + sin y + 1)i + (e x x cos y)j Ans. We have P (x, y) ye x + sin(y), Q(x, y) e x + x cos(y). Then Q x P y ex + cos(y). So Q P x y ex + cos(y) and F is conservative. To find f(x, y) such that F f, let f x f(x, y) P (x, y) and f y Q(x, y). Then (ye x + sin(y))dx ye x + x sin(y) + k 1 (y) ex + cos(y), and and f(x, y) (e x + x cos(y))dy ye x + x sin(y) + k 2 (x). From these equations, we have that k 1 (x) k 2 (y) C, where C is a constant. f(x, y) ye x + x sin(y) + C. When C, so f(x, y) ye x + x sin(y). Hence +2 : if you show that F is conservative well +2 : if you give the function f

10 STS26-4 The fourth exam, Page 1 of 1 Dec 15, Let F(x, y, z) yzi + xzj + (xy + 2z)k, C be the line segment from (1,, 2) to (4, 6, 3). (i) Find a function f such that F f and (ii) use part (i) evaluate F dr along the curve C. C Ans. (i) Since f f x i + f y j + f z k, then f x yz, f y xz and f z xy + 2z. Therefore f(x, y, z) f z dz xyz + z 2 + g(x, y). From f(x, y, z), f x yz + g(x,y) x f x yz, we get g(x,y) h(y) y. If we compare to xz then, that is, g(x, y) h(y). Also, since f x y xz + h(y) y, that is, g(x, y) h(y) c. Hence f(x, y, z) xyz + z 2 + c, where c is constant. (ii) By the fundamental theorem for line integrals, we have F dr f(4, 6, 3)f(1,, 2) ((2) 2 ) 77. C +2 : if you give the function f well +2 : if you find the line integral by using the fundamental theorem for line integrals