ln e 2s+2t σ(m) = 1 + h 2 x + h 2 yda = dA = 90 da R

Transcription

1 olution to et 5, Friday ay 7th ection 5.6: 15, 17. ection 5.7:, 5, 7, 16. (1) (ection 5.5, Problem ) Find a parametrization of the suface + y 9 between z and z. olution: cost, y sint and z s with t π and z. () (ection 5.5, Problem 4) Find a parametrization of hyperboloid of one sheet + y z 1 between z and z 1. olution: z s, 1 + s cos t and y 1 + s sin t for s 1 and t π. () (ection 5.5, Problem 6) Find parametrization of the plane + 6y + z 1 for in the first octant. We can choose t, y s and z 1 t 6s with t 1 6s, s. Alternately, we take obvious three points on the plane (1,, ), (,, ) and (,, 1). Vectors a (,, ) and b (,, 1) are in the plane of the the plane. o, parametric equation of the plane (, y, z) (1,, )+t(,, )+s(,, 1), with t 1, s 1 (4) (ection 5.5, Problem ) et up the double integral that represents the area of the parametrized surface ( e s+t, s t, s + t ), s 1, t ln olution: We have f s (e s+t,, 1) and f t (e s+t,, 1). We note that f s f t (,, e s+t ). o, we have f s f t 1 + e s+t. Therefore, area of the given surface is given by σ ln 1 + e s+t dsdt (5) (ection 5.5, Problem 4) Find the area of the surface, where is the portion of the plane z 8 + 5y + that lies above the region in the y-pane bounded by the parabola y and the line 4. olution: ince surface is given by z 8+5y+ h(, y), we have area of the surface σ() 1 + h + h yda dA 9 da Now is the region in the y plane with left curve y and right curve 4. Note that the curves intersect where 4 y, i.e. y ±. o, y ranges from - to in. o, from 1

2 above σ() 9 4 y d 9 9 ( 4 y ) dy (4 y )dy 9 ] 4y y 1 (6) (ection 5.6, Problem ) Evaluate the surface integral + y dσ, where is the cylindrical surface parametrized by ( coss, sin s, t) for s π, t 1. olution: We note that f s ( sin s, coss, ) and f t (,, 1). o, f s f t ( coss, sin s, ) and therefore, f s f t. Therefore, π π + y dσ 4 cos s + 4 sin s f s f t dsdt 4 dsdt 16π (7) (ection 5.6, Problem 4) Evaluate z + y dσ, where is parametrized by (θ cosθ, θ sin θ, w), θ 6π, w. olution: Here the parameters (s, t) have been replaced by (θ, w). We note that f θ (cosθ θ sin θ, sin θ + θ cosθ, ), f w (,, 1). o, o, f θ f w 1 f θ f w (sin θ + θ cosθ, cosθ + θ sin θ, ) sin θ + θ sin θ cosθ + θ cos θ + cos θ θ sin θ cosθ + θ sin θ 1 + θ We also have on, + y θ cos θ + θ sin θ θ θ (since θ ). o, z + y dσ w (1 + θ ) /] θ6π 6π θ dw 1 9 w θ 1 + θ dθdw (1 + 6π ) / 1 ] w ] w 16 w (1 + 6π ) / 1 ] 9 (8) (ection 5.6, Problem 6) alculate (y + z)dσ, where is the part of the plane + y + z that lies above the region { (, y) 1, y 1 } in the y-plane. olution: Note in the representation given the surface is conveniently decribed by z y h(, y). We have

3 1 + h + h y o, treating as a y-simple region in the y plane, / (y+z)dσ (y+h(, y)) 1 + h + h yda ( ) 6dyd 6 6 / / ( )y] y1 y d 6 ( ) d 6 / + 11 ( )(1 )d ] 1/ ] 18 (9) (ection 5.6, Problem 1) alculate F ndσ, where F (, z, y) and is the surface of revolution parametrized and oriented by (s, (s s) cost, (s s) sin t) f(s, t), s 1, t π. olution: Note f s (1, (s 1) cost, (s 1) sin t) and f t (, (s s) sin t, (s s) cost). We find f s f t ( (s 1)(s s), (s s) cost, (s s) sin t ) o, π π F ndσ π π F (f s f t )dsdt (, z, y) ((s 1)(s s), (s s) cost, (s s) sint ) dsdt s(s 1)(s s) + (s s) sin t cost ] dtds ts(s 1)(s s) + (s s) sin t ] tπ πs(s 1)(s s)ds 4π s(s s) s s5 s ] s1 s π t ds s(s 1)(s s)ds ] π (1) (ection 5.6, Problem 15) alculate F ndσ, where F (, y, z) and is the sphere parametrized and oriented by (, y, z) (sin φ cosθ, sin φ sinθ, cosφ), where θ π, φ π. olution: f φ (cos φ cosθ, cosφsin θ, sin φ), f θ ( sin φ sin θ, sin φ cosθ, ). We have f θ f φ ( sin φ cosθ, sin φ sinθ, cos φ sinφ )

4 4 From the statement about orientation, this is to be taken as the direction of n. Therefore, π π π π F ndσ (cosθ sin φ)(sin φ cosθ) + (sin φ sin θ)(sin φ sinθ) + cosφ(cosφsinφ) ] dφdθ sin φ + sin φ cos φ ] π dφdθ π sin φdφ π cosφ] π 4π (11) (ection 5.6, Problem 17) alculate F ndσ, where F (y,, z) and is the boundary of the cylindrical solid bounded by + y 4, z and z. Use the outward normal. olution: Note that the closed boundary of the solid is composed of three separate pieces: the bottom flat cap 1 : { (, y, z) : z, + y 4 } with outward normal n (,, ) top cap : { (, y, z) : z, + y 4 } with outward normal n (,, 1) and the cylindrical boundary {(, y, z) : (, y, z) ( cosθ, sin θ, z), θ π, z } We note that on, f θ ( sin θ, cosθ, ) and f z (,, 1) and so f θ f z ( cosθ, sin θ, ), which is indeed directed outwards from the cylinder. On 1 F n (y, z, z) (,, ) z. o, 1 F ndσ. On, F n (y, z, z) (,, 1) z. o, F ndσ da (π ) 8π +y 4 On, F ndσ π π F (f θ f z ) dθdz ( sin θ) ( cosθ) + (z )( sin θ) ] dθdz { 8 sin θ z cosθ ] θπ θ } dz o, we have total contribution F ndσ + 8π + 8π.

5 (1) (ection 5.7, Problem ) Evaluate the integral ddy by +y converting into polar coordinates where is the annular region 4 + y 9. olution: In polar coordinates (r, θ), region is described by r and θ π. We also have Jacobian (,y) r as (r,θ) worked out in class. o, we have ddy π + y rdrdθ r π ] dr dθ ( )(π) π (1) (ection 5.7, Problem 5) Evaluate the double integral arctan y da by converting into polar coordinates, where is the first qudrant between the circles y and + y 1 and the lines y and y. 5 y y/sqrt() r1/ r1 Figure 1. egion between y, y /, + y 1 4 and + y 1 olution: Note integrand arctan y θ in polar coordinates. ince y in polar coordinates implies tanθ 1, we have θ π since we are in the first quadrant. Again y 4 implies tan θ 1, so θ π since we are in the first quadrant. Therefore, the description of given region in polar coordinates 6 is { (r, θ) : 1 r 1, π 6 θ π } 4

6 6 o, recalling from class Jacobian (,y) (r,θ) arctan y da π/4 π/4 π/6 r θ ] r1 π/6 r1/ dθ 8 1 π/4 π/6 θrdrdθ r, we have θdθ ] θ θπ/4 16 θπ/6 15π (16)(144) (14) (ection 5.7, Problem 7) Using polar coordinates evaluate the double integral 4 y y 4 y ddy. e y sqrt(4 y ) sqrt(4 y ) y Figure. egion between 4 y and 4 y for y. learly a circle of radius centered at (, ) olution: From the order of integration, it is clear that the above is the area integral over region treated as an -simple region between 4 y amd 4 y for. This is clearly the interior of the circle +y 4. o, polar representation of domain is {(r, θ), r, θ π}. o, it follows that 4 y y e 4 y π ddy e r rdrdθ e r da π 1 e r ] r dθ π 1 e 4] r (15) (ection 5.7, Problem 16) Evaluate the double integral sin(y) ddy y by making an appropriate change of variables where is bounded by πy πy, y and y. We choose s and y

7 7 t y. Then the domain in the s t variable becomes {(s, t) : π s π, 1 } t 1 Also, note that s t and st y, so s / t 1/, y s 1/ t / and we obtain (, y) J (s, t) sy t y s t Therefore, noting that the integrand y obtain sin(y) ddy 1 y 1 π π/ π π/ 1 π cos(st)] t1 t1/ ds π/ 1 sin s + sin s ] sπ s sin(st)] dtds s π sin(y) s sin(st), we 1 coss + 1 cos s ] ds (16) (ection 5.8, Problem ) By converting into cylindrical coordinates, evaluate 1 y zdzdyd 1 z z y y Figure. egion between z and z y whose projection on the y plane is the unit semi-circle centered at the origin for which

8 8 olution: We note from the order of the integration that the solid that corresponds to the volume integral above is a z- simple with an upper surface at z y h (, y), which is a paraboloid facing down and a lower-surface z h 1 (, y), which is a plane parallel to the y plane. The projection of the solid region in the y plane, call it is being treated as a y simple region with lower curve y 1 and upper-curve y 1, each being on the circle + y 1. However, is not the full circle since from the outermost limit,. Hence is only the semicircle {(, y) : + y 1, }. We note that in polar coordinates the description of is { (r, θ) : r 1, π θ π}. o, is the region between the two surfaces z h 1 (, y) and z h (, y) whose projection on the y plane is the semicircle. It is useful to sketch as given above. onverting into cylindrical coordinates and doing the w (or z) integration first, we have y zdzdyd zdv 1 1 π/ r π/ π/ π/ wrdwdrdθ π/ π/ 1 r( r ) r ] dr π ] w r 1 w r drdθ w 16 ] r1 ( r ) r r π ] 4 π (17) (ection 5.8, Problem 5) Using cylindrical coordinates calculate 1/ 1 +y ( + y )dzdyd / +y +y olution: Looking at the limits, we have 1 +y ( + y )dzdyd ( + y )dv where is being treated as a z-simple region in -D going from the paraboloid z + y 1 to the cone z + y. o, we may write r ] ( + y )dv r dz da r

9 The projected region in the y plane is obtained by looking at the two outer integrals. ince y integration is done first, the region is being described by {(, y) : y } 1, / Note that the upper and lower curves intersect when 1, i.e. 1/sqrt. We plot this as shown below: y y 9 ( /, ) / Figure 4. egion between curves y and y 1 for / ince the part of the straightline y for > in the y plane corresponds to θ π, while y 1 4 corresponds to + y 1 1 r, i.e. r 1, we have the description of region in the (r, θ) domain as { (r, θ) : r 1, π 4 θ π } Therefore, since Jacobian J r, it follows that π/ r ( + y )dv r (r)dwdrdθ π/ r π/4 r π/4 r r w ] π/ wr drdθ r (r r +1)drdθ π wr π/4 4 π ( ) π (18) (ection 5.8, Problem 8) Using spherical coordinates, evaluate + y + z ddydz, ( r 4 r 5 + r ) dr

10 1 where is the solid in the first octant bounded by the planes y, y, the cone z + y, the plane z and the spheres + y + z and + y + z 8. oluton: Note that the description of the spheres in spherical coordinates is ρ and ρ 8. ince tan θ y, the description of the part of the plane in the first octant y is θ π and 4 the plane y is θ π, Further, since z ρ cosφ, it follows that z plane corresponds to φ π. Again, z + y implies ρ cos φ ρ sin φ cos θ + ρ sin φ cos θ ρ sin φ. o, z + y implies φ π. o, the region in spherical 4 coordinate description is { (ρ, θ, φ) : ρ 8, π 4 θ π, π 4 φ π } We recall the absolute value of the Jacobian J between the transformation from cartesian to spherical coordinates is ρ sin φ o, (8/ / ) π/ + y + z ddydz π/ π/ π/4 π/4 π/ π/ π/4 π/4 π/4 1 ρ cosθ sin φ (8/ / ) π/ 8 π/4 ] ρ 8 dθdφ ρ cosθ sin φdθdφ (8/ / ) ( 1 ρ sin φ cosθ (ρ sin φ)dρdθdφ ρ π/ π/4 ) π/ sin φdφ ince sin φ (1 cos φ)dφ φ 1 sin(φ)], we obtain 4 answer ( ) (π (8 / / ) ) 4 (19) (ection 5.8, Problem 11) Using spherical coordinates evaluate π/4 sin θ sin φ ] θπ/ θπ/4 dφ 9 9 y ydzdyd

11 olution: Looking at the order of integration and limits we have 9 y ] 9 z ydzdyd ydz da 9 where is the projected region in the y plane between which is given by the description { (, y) : y } 9, This is clearly the first quadrant in the interior of the circle of radius centered at the origin. o, the triple integral represents the volume integral ydv, where is described as the -D region whose projection on the y plane is and lies between z and z 9 y, i.e. in the first octant of the sphere of radius, centered at the origin. Therefore, recalling that Jacobian J ρ sin φ, we obtain π/ π/ ydv π/ π/ π/ 1 5 ρ5 sin φ cosθ sin θ π/ 1 sin φ sin θ 11 ρ sin φ cosθ sin θ ] (ρ sin φ)dρdθdφ ] ρ dθdφ 4 ρ 5 ] θπ/ θ 1 cos φ ] dcosφ] 4 1 dφ 4 1 π/ π/ π/ sin φdφ cosφ 1 cos φ sin φ cosθ sin θdθdφ ] φπ/ φ 81 5 () (ection 5.8, Problem 1) By making appropriate changes of variables, calculate ( + z) ddydz, where is the solid bounded by the planes +y +z, +y +z 1, y +z, y + z 1, + z and + z 1. olution: We choose s +y+z, t y+z, u +z. o the image of region under the transformation (, y, z) (s, t, u) is the region {(s, t, u) : s 1, t 1, u 1} Now note (s, t, u) (, y, z) det

12 1 o, needed jacobian J (,y,z) 1/() and J 1. (s,t,u) o, ( + z) dv u (1)dsdtdu u dsdtdu u s ] ] 1 u dtdu u 1 dtdu s u ( ) (1) (ection 6.1, Problem ) Determine if F(, y) y + 1, y 1 y is conservative in D {(, y) y > }. olution: Note F 1 (, y) y +1, F (, y) y 1. We y also note F 1 y 6y F Further the function F is continuously differentiable in D since domain D does not contain y where the function is not continous/differentiable. Therefore, from theorem given in class, F is conservative. () ((ection 6.1, Problem ) 4) Determine if F(, y, z)(f 1, F, F ) ln y, z, 1 is conservative in the domain D {(, y, z) y > }. y olution: We note that i j k F y z ln y z 1 y ( 1 ] y y z (z), ( ln y) 1 ], z y (z) ) ( ln y) y ( ) 1 y 1,, y Hence F is not conservative. () (ection 6.1, Problem 16) alculate y5/ d+ 5 y/ dy, where is the parabolic path y from (, ) to (, 8). olution: First, check if F (F 1, F ) ( y 5/, 5y/) is conservative or not. We note F 1 5 y y/ F. o, F is conservative and therefore from Theorem in class, there eists scalar potential f so that f F. Now we calculate f. ince f (f, f y ) (F 1, F ), f y 5/, and f y 5 y/

13 Partially integrating the first relation, we have f(, y) y 5/ + g(y). ubstituting into second equation, we have 5 y/ + g (y) 5 y/ Therefore, g (y) and so g(y). o, f(, y) y 5/ +. o, with F defined above, we note (.1) y 5/ d+ 5 y/ dy F d ( f) d f(, 8) f(, ) (8) 5/ 64 The answer does not depend on the path since F was checked to be conservative (path-independent). (4) (ection 6.1, Problem 18) alculate yd dy, where is the circle centered at the origin traversed counter-clockwise once. olution. We define F(, y) (y, ) (F 1, F ) noting that yd dy F d. We first note that F 1 F 1 ; so F is not conservative and we do not have a short cut to get to the answer. We have to use the basic line integral definition and start first with parametrization of path : (cost, sin t) for t π in order to traverse the unit circle counterclockwise once. o, yd dy π sin t( sin tdt) cos t(costdt)] π 1 ]dt π (5) (ection 6.1, Problem ) Determine ye dy+e dy+dz where is the line segment from (ln,, 4) to (, 1, ). olution Let s check first, if the F (ye, e, 1) is conservative. We note that F i j k y z ye e 1 ( (1) y (e z, (ye ) (1) ) z, (e ) (ye (,, ) y o, F is conservative and the answer is not dependent on the path but only on the end points. We also have some scalar potential f so that F (ye, e, 1) f (f, f y, f z ). o, f ye, f y e, f z 1 o, partially integrating first epression, f(, y, z) ye + g(y, z). ubstituting into second epression f y e +g y (y, z) e. o, g y (y, z) implying g(y, z) h(z) and we have so

14 14 far f(, y, z) ye + h(z). ubsituting into third epression, f z h (z) 1. o, h(z) z +. o, we have scalar potential f(, y, z) ye + z + Therefore, using theorems in class ye dy + e dy + dz F d ( f) d f(, 1, ) f(ln,, 4) e + e ln ( 4) (6) (ection 6.1, Problem ). alculate yzd+ zdy+ ydz where is the closed contour along the intersection of the sphere + y + z 1 and the plane z 1. olution: First check if F (yz, z, y) is conservative, because if this is true, than according to the theorem in class, F d and we don t have to do any line integral calculations. We calculate F i j k y z yz z y ( ( y) ( z), (yz) ( y) y z z, ( z) (yz) ) y (, y y, z z) o, F. Therefore, F is conservative and therefore from theorem in class the closed path integral yzd + zdy + ydz F d