Tufts Univesit Math 3 Depatment of Mathematics Novembe, Eam : noon to : pm Instuctions: No calculatos, notes o books ae allowed. Unless othewise stated, ou must show all wok to eceive full cedit. Simplif ou answes as much as possible. Please cicle ou answes and coss out an wok ou do not want gaded. You ae equied to sign ou eam book. ith ou signatue ou ae pledging that ou have neithe given no eceived assistance on the eam. Students found violating this pledge will eceive an F in the couse.. ( points) Tue o False - No Patial edit: On the fist page of ou blue book, answe the following questions as Tue o False. (a) + d d + d d Solution: False: the outside integal on the ight has an in its limits, which isn t allowed. (b) The point (,, z) (, 3, ) in atesian coodinates is the same as 3, θ 6, z 6 in clindical coodinates. Solution: False: the z-coodinates of the two points ae diffeent. (c) The point (,, z) (,, ) in atesian coodinates is the same as ρ, ϕ, and θ in spheical coodinates. Solution: Tue: Fist note that ϕ means that the point is in the plane, so has coodinate z. Then compute fom the fomulas: ρ cos θ sinϕ, ρ sinθ sinϕ. (d) The vecto field F(,, z),, is consevative. Solution: Tue: a potential function fo it is f(,, z). (e) If F is a consevative field and is paameteized b (t), a t b, then F d F((b)) F((a)). Solution: False: The integal on the left should give a scala as an answe, since it is the line integal of a vecto field. The values on the ight, howeve, ae vectos.. ( points) (a) Epess the volume of the solid egion, E, that sits above the ectangle in the -plane with vetices (,, ), (,, ), (,, ), and (,, ) and below the suface z in tems of a double integal. Solution: The volume of the egion between z and z within the ectangle [, ] [, ] in the plane is given b the double integal dd. (b) Evaluate the integal ou found in pat (a). Solution: dd d 3 d 3 5.
3. (5 points) Let R be the egion in the plane inside the cicle + and above the line. (a) Sketch R. Solution: The cicle + can be ewitten as + ( ), so it is the cicle of adius centeed at (, ). The uppe half of this cicle is above the line. R (b) Epess R in pola coodinates. Solution: Note that we can ewite the line as sin θ, o sin θ. e can also ewite the cicle + as sinθ, o sin θ. Thus, we can wite both the top and bottom boundaies of R b giving in tems of θ. e now need to find limits on θ. To do this, notice that the limits of the egion in θ will occu when sin θ and sinθ ae both satisfied: sin θ sinθ, which gives sin θ, o θ and θ 3. (Thee ae, of couse, othe values of θ that satisf sin θ, but the egion, R, onl etends into the fist and second quadants. Putting this togethe gives R { (, θ) θ 3, sin θ sinθ}. (c) ompute R f(, )da fo f(, ) +.. ( points) Solution: Fist, we make the change of vaiables fom (, ) to (, θ): R f(, )da sin θ sin θ Integating with espect to, we get R f(, )da sin θ sin θ cos θ ddθ cos θdθ sin θ sin θ cos θddθ. ( sin θ ) sin cos θdθ. θ To evaluate this, notice that we can easil make a u-substitution fo u sinθ, giving du cos θdθ, and f(, )da u R udu. (a) Rewite the iteated integal dz d d
as an iteated integal in the ode dz d d. Solution: e onl need to echange the outside ode in and, so we don t need to wo about the limits in z. So, we daw the sketch of the egion, D, in the plane: D Fom the pictue, we notice that the maimum limits in ae and while, fo fied, vaies fom to. This gives dz d d dz d d (b) Evaluate one of the integals fom pat (a). Solution: Eithe integal seems easonable to compute. Using the second one, we get dz d d o ( )dd dd 3 3 d 3 3 d 5 5 5 6 5 8 3 6 8 3. 5. (5 points) Let f(,, z) z and let be the egion that is above the cone z + and below the sphee + + z 8. (a) Setup (but do not evaluate) the integal f(,, z)dv in both i. lindical oodinates, and Solution: In clindical coodinates, the cone z + becomes z, while the sphee becomes + z 8. These two sufaces intesect when + 8, o. This gives limits on the z integal as z 8 (since is above the cone and below the sphee), on the integal as (because the maimum etents in the -plane coespond to a cicle of adius ), and on the θ integal as θ (since thee ae no estictions on θ. Thus, 8 z dv zdzddθ.
ii. Spheical oodinates. Solution: In spheical coodinates, the sphee is given b ρ 8, o ρ q. The cone is given b z + ρ cos ϕ ( ρ cos θ sin ϕ + ρ sin θ sin ϕ ) ρ sinϕ. So, cos ϕ sinϕ o tanϕ. Above (inside) the cone coesponds to angles smalle than, so the limits on ϕ ae ϕ. The cone gives no limits on ρ, so the limits on ρ ae given b ρ. No limits ae given on θ, so θ. Thus, z dv ρ cos ϕρ sin ϕdρdϕdθ ρ 3 cos ϕ sinϕdρdϕdθ. (b) Evaluate f(,, z)dv using one of the integals in pat (a). Solution: Eithe integal is eas to evaluate. In clindical coodinates: z dv In spheical coodinates: z dv 8. 6. ( points) Evaluate V 8 8. z z zdzddθ z 8 ρ ρ dθ ddθ ρ 3 cos ϕ sin ϕdρdϕdθ. ρ 8 8 sin ϕ ϕ ϕ dθ ( + ) / dv, whee cos ϕ sinϕdϕdθ dθ (6 8)dθ V {(, θ, z) 6, θ, z /}. 8 3 ddθ 6 cos ϕ sinϕdϕdθ
Solution: Using the clindical coodinates desciption of V given to us, we have V ( + dv ) / 6 / 6 sinθddθ sinθ dzddθ sinθdθ cos θ ( ) ( ). 6 / sin θdzddθ 7. ( points) Evaluate the line integal (3 )ds whee is the cuve paameteized b (t) t, 3t fo t. Solution: Fom the definition, f(, )ds omputing (t), 3, so vec (t) 5, we have (3 )ds 8. ( points) Let f(,, z) cos()sin()e z. (a) ompute the gadient field, F f. Solution: B the definition, F f f, f, f z f(t, 3t) (t) dt. ( 3(t) (3t) ) 5dt 5(8t 8t )dt 5t 3. 5t dt sin()sin()e z, cos()cos()e z, cos()sin()e z. (b) ompute the integal, F d, whee is the path given b (t) cos(t), sin(t), cos(t), t. Solution: Since F is defined as the gadient of f(,, z), we know that F is consevative. The integal of an continuous consevative field ove a closed path is alwas zeo, so F d. (Note that is closed: () ().) 9. ( points) Let F(,, z) + e z, sin(), e z + 7. Is F consevative? If so, find a potential function fo F.
Solution: e fist check that F is consevative. iting f(,, z) + e z, g(,, z) sin(), and h(,, z) e z + 7, we have f g f z ez h g z h. These thee conditions veif that F is consevative. To find a potential function, φ(,, z), stat b noticing that φ f(,, z) + ez. So φ(,, z) + e z d + e z + (, z). Diffeentiating this with espect to gives φ +. Matching this with g(,, z) gives ( (, z) sin() ) d cos() + K(z), o φ(,, z) + e z + cos() + K(z). Diffeentiating this with espect to z gives φ z ez + dk dz. Matching this with h(,, z) gives K(z) (e z + 7) e z dz 7z + K. Finall, this gives φ(,, z) + e z + cos() + 7z + K. To veif that φ is a potential function fo F, compute φ + e z, sin(), e z + 7. End of Eam