(0,2) L 1 L 2 R (-1,0) (2,0) MA4006: Exercise Sheet 3: Solutions. 1. Evaluate the integral R

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1 MA6: Eercise Sheet 3: Solutions 1. Evaluate the integral d d over the triangle with vertices ( 1, ), (, 2) and (2, ). Solution. See Figure 1. Let be the inner variable and the outer variable. we need the equations of L 1 and L 2. Use m + c and substitute in points ( 1, ), (, 2). We nd that m c 2 and so or ( 2)/2. Similarl, L 2 is the line + 2 and so + 2. Hence d d 2 +2 ( 2)/2 d d Note that this is precisel the area of the triangle. 2 ( 32 ) + 3 d 3. (,2) L 1 L 2 (-1,) (2,) Figure 1: Question Evaluate the integral 1/2 d d over the area bounded b 2, + 2, and the -ais. Solution. See Figure 2. Let be the inner variable and the outer variable. Then 1/2 d d [2 1/2] 2 1/2 d d 2 d ( 2(2 ) 1/2 2 ) d [ 3 (2 )3/2 2 ] Evaluate Ω ds where Ω(, ) and is the segment of the line 3 from (, ) to (2, 6). Solution. Let t and then 3 3t. The point (, ) is t and the point (2, 6) is 1

2 (,2) 2 (1,1) (2,) Figure 2: Question 2. t 2. Ω ds 2 Ω(r(t)) r (t) dt. Now, r(t) (t, 3t) and so r (t) (1, 3), r (t) 1. Also Ω(r(t)) t 2 + 9t 2 1t 2. So Ω ds 2 1t 2 1 dt Evaluate f ds where f(,, z) z 2 and : r (t, cos t, sin t), t 2π. Solution. Here r (t) (1, sin t, cos t) and so r (t) 2. Also f(r(t)) 1 + cos 2 t + sin 2 t 2. f ds f(r(t)) r (t) dt 2 2 dt 2π. 5. Find the work done in moving a particle in a force eld given b F 3i 2 j + 1k along the curve 2 2, z, from the point (,, ) to the point (1, 2, ). Solution. The work done is F dr. To nd the parametric denition of, we set t and then 2 2 2t 2, with z. The points (,, ) and (1, 1, ) correspond to t and t 1 respectivel. So r(t) (t, 2t 2, ), t 1, giving r (t) (1, t, ) and F(r(t)) ( 3(t)(2t 2 ), (2t 2 ) 2, 1(t) ) (6t 3, t, 1t). 2

3 F dr F(r(t)) r (t) dt (6t 3, t, 1t) (1, t, ) dt (6t t 5 ) dt If f 8 2 zi + 5zj k, nd the work done in moving a particle in a force eld given b f along the curve with parametric denition r(t) (t, t 2, t 3 ), ( t 1). Solution. Here r (t) (1, 2t, 3t 2 ) and f(r(t)) ( 8(t) 2 (t 2 )(t 3 ), 5(t 3 ), (t)(t 2 ) ) (8t 7, 5t 3, t 3 )., f dr f(r(t)) r (t) dt (8t 7, 5t 3, t 3 ) (1, 2t, 3t 2 ) dt (8t 7 +1t 12t 5 ) dt Show that f 2 i + 2j is a conservative vector eld. Determine and associated scalar potential ϕ for this vector eld f. Solution. f i j k 2 2 i( ) j( ) + k(2 2). Hence f is conservative and so there eists ϕ such that f ϕ, i.e. ( 2, 2, ) ( ϕ, ϕ, ϕ ), or (i) ϕ 2, (ii) ϕ 2, (iii) ϕ. (i) ϕ 2 and so ϕ 2 + g(, z) ϕ 2 + g. (1) (ii) Also ϕ ϕ 2 + g(z). g 2 and so comparing with (1) we see that. So g g(z) and 3

4 (iii) Finall, ϕ g which means that and so g is constant. the scalar potential is ϕ(,, z) 2 + c where c is an arbitrar constant. 8. If a ( , 1z, 2z 2 ) evaluate a dr from (,, ) to (1, 1, 1) along the following paths: (a) The straight lines from (,, ) to (1,, ), then to (1, 1, ) and nall to (1, 1, 1). (b) The straight line from (,, ) to (1, 1, 1). Solution (a). The line from (,, ) to (1,, ) is parameterised b 1 : r(t) (t,, ), t 1. a(r(t)) (3t 2,, ) & r (t) (1,, ), and so I 1 a(r(t)) r (t) dt The line from (1,, ) to (1, 1, ) is parameterised b 3t 2 dt 1. 2 : r(t) (1, t, ), t 1. a(r(t)) (3 + 6t,, ) & r (t) (, 1, ), and so I 2 a(r(t)) r (t) dt Finall, the line from (1, 1, ) to (1, 1, 1) is parameterised b dt. 3 : r(t) (1, 1, t), t 1. and so a(r(t)) (9, 1t, 2t 2 ) & r (t) (,, 1), I 3 Hence a dr I 1 + I 2 + I a(r(t)) r (t) dt 2t 2 dt 2 3.

5 Solution (b). The line from (,, ) to (1, 1, 1) is parameterised b : r(t) (t, t, t), t 1. a(r(t)) (3t 2 + 6t, 1t 2, 2t 3 ) & r (t) (1, 1, 1), and so a dr a(r(t)) r (t) dt (3t 2 + 6t 1t 2 + 2t 3 ) dt Let ϕ and let V denote the closed region bounded b the planes z 8,, and z. Evaluate V ϕ dv. Solution. See Figure 3. If z is the inner variable then z varies from to z z z plane (setting z ) 2 Figure 3: Question 9. If is the middle variable then (in the -plane, setting z, see Figure 3) varies from to 2. If is the outer variable then varies from to 2. V ϕ dv dz d d (8 2) d d ( ) dz d d ( ) d

6 1. Using Green's Theorem, evaluate the line integral F dr counterclockwise around the boundar of of the region, where f (sin, cos ) and is the triangle with vertices (, ), (π, ) and (π, 1). Solution. Green's Theorem sas F dr Now r (sin, cos ) and f 1 cos and f 2 are shown in Figure. 1 ( f2 f ) 1 d d. sin. The egion and boundar curve / Figure : Question 1. F dr ( sin cos ) d d π /π π π ( sin cos ) d d [ ] /π sin sin d ( π sin sin ) π d π cos 1 1 π. 11. Evaluate S a ˆn ds S a ds where a (18z, 12, 3) and S is part of the plane z 12 located in the rst octant. Solution. See Figure 5. Plane z 12 is z g(, ). Let r(u, v) r(, ) (,, g(, )). So r (1,, g ) (1,, 1/3), r (, 1, g ) (, 1, 1/2). 6

7 z z 12-2/ Figure 5: Question 11. r r i j k 1 1/3 1 1/2 (1/3, 1/2, 1). Also a(r(, )) (36 6 9, 12, 3). So a ds S a (r r ) d d 6 2/3 6 6 (36 6 9, 12, 3) (1/3, 1/2, 1) d d [ ] 2/3 (6 2) d ( ) d Use the Divergence Theorem to compute the surface integral S a ds where a (, 2 2, z 2 ) and S is the clindrical surface 2 + 2, z 3. Solution. The Divergence Theorem states that S a ds V a dv. Now ( a,, ) (, 2 2, z 2 ) + 2z. Use clindrical co-ordinates r (r cos θ, r sin θ, z), r 2, θ 2π, z 3. 7

8 Then J V (,, z) (r, θ, z) a dv r, ( a)(r(r, θ, z)) r sin θ + 2z ( r sin θ + 2z)r dr dθ dz (r r 2 sin θ + 2zr) dr dθ dz (8 323 ) sin θ + z dθ dz (16π + 8πz) dz 8π. 13. Verif Stokes' theorem for the case f (2, z 2, 2 z) where the surface S is the hemisphere z 2 1, z and the curve is dened as the unit circle. Solution. See Figure 6. Stokes' Theorem: f dr S ( f) ds. First let us nd z Figure 6: Question 11. f dr. Since is the unit circle we have r(t) (cos t, sin t, ), t 2π. Then 8

9 r (t) ( sin t, cos t, ) and f(r(t)) (2 cos t sin t,, ). f dr + f(r(t)) r (t) dt (2 cos t sin t,, ) ( sin t, cos t, ) dt ( 2 sin t cos t + sin 2 t) dt ( sin 2t + sin 2 t) dt 1 (1 cos 2t) dt π. 2 Now nd S ( f) ds. We have i j k f 2 z 2 2 z (,, 1). The surface of hemisphere is given parametricall b r(θ, ϕ) (sin θ cos ϕ, sin θ sin ϕ, cos θ), θ π/2, ϕ 2π, and r θ (cos θ cos ϕ, cos θ sin ϕ, sin θ) r ϕ ( sin θ sin ϕ, sin θ cos ϕ, ) r θ r ϕ (sin 2 θ cos ϕ, sin 2 θ sin ϕ, sin θ cos θ). Hence S ( f) ds π/2 π/2 π/2 (,, 1) (r θ r ϕ ) dθ dϕ sin θ cos θ dθ dϕ dϕ π. sin 2θ dθ dϕ We have therefore veried Stokes' Theorem. 9