SAMPLE MATH 209 FINAL EXAM SOLUTIONS. James D. Lewis. [ ] y sin x

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1 1. Evaluate the integral AMPLE MATH 9 FINAL EXAM OLUTION James D. Lewis 1 1 y sin x x dxdy. olution: The integral in its present form is hard to compute. By graphing the region and reversing the order of integration, we arrive at the simpler integral 1 1 y sin x x dxdy 1 x sin x x dydx 1 [ ] y sin x yx dx x y 1 sin x dx 1 cos 1.. A thin plate D of uniform thickness covers the region enclosed by y x and y x 3. The density at a point in D is the distance to the line x 1. Find the centre of mass. olution: The region D is described by x 3 y x, x 1. For a point P (x, y) D, the density function is given by ρ(x, y) 1 x. We first compute the total mass: M 1 x x 3 (1 x) dydx 1 [ y(1 x) ] yx yx 3 dx 1 (x x 3 + x 4 ) dx Next, M x Finally 1 x M y 1 [ ] y yx y(1 x) dydx (1 x) dx 1 x 3 yx 3 1 [ ] x x 3 x(1 x) dydx (x, y) 1 [ y(x x ) ] yx yx 3 dx ( My M, M ) ( x 1 M, 13 ) (x 4 x 5 x 6 + x 7 ) dx (x 3 x 4 + x 5 ) dx

2 3. Let Ω be the solid bounded above and below by z x + y and on the sides by x + y + z 1. (i) ompute the volume of Ω using spherical coordinates. (ii) ompute the volume of Ω using cylindrical coordinates. [Hint: You may have to express Ω as a union of two solids.] olution: (i) By spherical, we have Vol(Ω) π 3π 4 π 4 1 ρ sin φ dρdφdθ π. 3 (ii) By cylindrical, we have Ω Ω 1 + Ω, where Ω 1 : 1 r z 1 r, 1 r 1, θ π. Ω : r z r, r 1, θ π. This uses the fact that the surfaces z x + y r (double cone with vertex (,, )) and r +z x +y +z 1 (sphere of radius 1 centred at (,, )), meet when r 1/. Thus we have Vol(Ω) Vol(Ω 1 ) + Vol(Ω ). We compute: Vol(Ω 1 ) π 1 1 r 1 1 r rdzdrdθ π 1 r 1 r drdθ π 1 3. Next Vol(Ω ) π 1 r r rdzdrdθ π 1 r drdθ π 3. Finally Vol(Ω) π /3. 4. Let be the path below with given orientation, and P x + y and Q y + x. : (,1) (1,1) (,) yx (i) Evaluate two ways, namely, P dx + Qdy

3 (i.1) Directly from the definition of line integrals. (i.) Using Green s theorem. (ii) Is the vector field given by F P i + Qj conservative? [You must give a reason for your answer.] olution. (i.1) Write as in the diagram below. We have these parametrizations: (,1) (1,1) 3 (,) 1 r(t) (t, t) : [, 1] 1, r(t) (t, 1) : [, 1], r(t) (, t) : [, 1] 3, where j is the the curve j with the opposite orientation. Thus P dx + Qdy 3 P dx + Qdy P dx + Qdy j 1 j1 P dx + Qdy 3 P dx + Qdy 1 (t + t, t + t) (1, 1) dt 1 ( ) (t +, 1 + t) (1, ) dt ( ) (i.) Let D be the region bounded by the curve. By Green s theorem P dx + Qdy D 1 (t, t ) (, 1) dt ( Q x P ) da da Area(D) 1 y D. (ii) The vector field F is clearly not conservative since rot(f) : Q x P y (a) Let be the curve given by r(t) (cos t)i + (sin t)j + (sin t)k with t π. 3

4 Find directly from the definition of line integrals. zdx + xdy + y dz (b) Let f(x, y) xy. ompute the path integral olution: (a) zdx + xdy + y dz π π fds. sin t d(cos t) + cos t d(sin t) + sin t d(sin t) ( sin t + cos t cos t sin t ) dt π. [Note the identities that were used, and that all students should know: sin t 1 ( ) 1 cos t ; cos t 1 ( ) 1 + cos t.] Next (b) r (t) ( sin t)i + (cos t)j + (cos t)k r (t) 1 + cos t. f(r(t)) cos t sin t, and observe that d(1 + cos t) cos t sin t dt. Thus by substitution (or an anti-symmetry argument), we arrive at fds 6. Evaluate the surface integral π cos t sin t 1 + cos t dt. F d, where F xi+yj+3zk, where is the cube with vertices (±1, ±1, ±1), with outward orientation. 4

5 olution: Method #1 (From the definition, and the less favorable method!) Write 6 j1 j, the six sides of the cube, with parametrizations Φ j : [ 1, 1] j, Φ 1 (y, z) (1, y, z) Φ (y, z) ( 1, y, z) Φ 3 (x, y) (x, y, 1) Φ 4 (x, y) (x, y, 1) Φ 5 (x, z) (x, 1, z) Φ 6 (x, z) (x, 1, z) Then correspondingly is the table of calculations: Thus F d n 1 (1,, ) F(Φ 1 ) (1, y, 3z) F n 1 1 n ( 1,, ) F(Φ ) ( 1, y, 3z) F n 1 n 3 (,, 1) F(Φ 3 ) (x, y, 3) F n 3 3 n 4 (,, 1) F(Φ 4 ) (x, y, 3) F n 4 3 n 5 (, 1, ) F(Φ 5 ) (x,, 3z) F n 5 n 6 (, 1, ) F(Φ 6 ) (x,, 3z) F n 6 6 j1 j F d 6 j1 j (F n j )d 1 Area ( [ 1, 1] ) 48. Method # (uperior method!): Use the divergence theorem. Let Ω [ 1, 1] 3 be the cube with boundary. Then div(f) 6, hence F d (F n)d div(f) 6Vol(Ω) Any of the questions below could serve as a question 7: (i) Verify the Divergence Theorem F nd Ω E div(f)dv, where F (xz)i+(yz)j+(3z )k, and E is the solid bounded by the paraboloid z x +y and the plane z 1. olution: div(f) F 8z. Thus E π 1 1 div(f)dv 8 rzdzdrdθ 8π r 3. On the other hand, the boundary of E is of the form 1 +, where 1 is the graph of z x + y over D : {x + y 1}, and 1 is the graph of z 1 over D. Note that the 5

6 unit normal for is k, and that F (x, y, 3) on, and hence F k 3 on. Therefore F d 3 D da 3π. If we put Φ(x, y) (x, y, x + y ) : D 1, then the unit outward on 1 is Further, Thus Φ x Φ y Φ x Φ y (x, y, 1) (x, y, 1). F(Φ(x, y)) (x 3 + xy, yx + y 3, 3(x + y ) ), F (Φ x Φ y ) (x + y ). 1 F d π 1 r 5 drdθ π 3. Finally, we arrive at F d F d + F d π π 8π 3. (ii) Verify tokes Theorem, url(f) d F dr, where F (3y)i + (4z)j + ( 6x)k, and is the part of the paraboloid z 9 x y that lies above the xy-plane, oriented upward. pecial Remark. Note that url(f) d ( url(f) n ) d, where n is the unit normal determined by the orientation on, and ( ) F dr F T ds, where T is the unit tangent determined by the (induced) orientation on. Thus tokes theorem is given by the equivalent statement ( ) ( ) url(f) n d F T ds. olution to (ii): Let D {x + y 9}. The boundary of is easily seen to be the same as the boundary of D, namely {x + y 9}, oriented counterclockwise. Put Φ(x, y) (x, y, 9 x y ) : D. 6

7 Then Φ x Φ y (x, y, 1) (note that this is oriented upward, since the z-component is 1 > ). Put r(t) (3 cos t, 3 sin t, ) : [, π]. Then F(r(t)) (9 sin t,, 18 cos t), and r (t) ( 3 sin t, 3 cos t, ). Thus F r (t) 7 sin t. Thus omputing the other way, we have π F dr 7 sin tdt 7π. url(f) F ( 4, 6, 3) ; url(f) (Φ x Φ y ) 8x + 1y 3. Thus url(f) d π 3 D ( 8x + 1y 3)dA ( 8r cos θ + 1r sin θ 3r)drdθ 7π. (iii) Use tokes theorem to evaluate url(f) d, where F (y z)i + (xz)j + (x y )k, and where is the part of the paraboloid z x + y inside the cylinder x + y 1. olution: If we set D {x + y 1}, then is given by the parametrization: Φ(x, y) (x, y, x + y ) : D. The boundary of D is parametrized by (cos t, sin t), and hence the boundary of is parametrized by r(t) : Φ(cos t, sin t) (cos t, sin t, 1); moreover F(r(t)) (sin t, cos t, cos t sin t) ; r (t) ( sin t, cos t, ). Thus url(f) d π F(r(t)) dr π ( (cos t 1) sin t + cos t ) dt π. ( sin 3 t + cos t)dt (iv) Use the divergence theorem to evaluate the integral: (x + y + z )d, 7

8 where is the sphere x + y + z 1. olution: The unit outward normal on is given by n (x, y, z). [This follows from n F/ F, where F x + y + z 1 describes implicitly.] Now choose F (,, z). Then F n x + y + z, and div(f) 1. Let Ω be the ball of radius 1. Then by the divergence theorem: (x + y + z )d Ω F nd dv Volume(Ω) 4π 3. Ω div(f)dv 8