Eample: 3.1 Find the area in the positive quadrant bounded b 1 and 3 4 First find the points of intersection of the two curves: clearl the curves intersect at (, ) and at 1 4 3 1, 1 8 Select a strip at an arbitrar location of. The area of the strip is given as A ( ) 1 4 3 (this is the upper curve minus the lower curve) 1/4 (1/, 1/8) (1/, 1/8) 4 3 1/3 Now take the limit d, and add up all the A strips A 1/ ( ) 1 4 3 d ( ) 1 4 4 4 1/ 1 64 For some functions, it ma be more convenient to sum up strips. Rearrange the equations of the curves in terms of 3 or 1/3 1
and 1 or 4 4 Select a strip at an arbitrar location of. The area of the strip is given as A ( 1/3 4 ) and the total area is A 1/8 ( 1/3 4 ) d ( ) 3 4 4/3 4 1/8 1 64
Eample: 3. Find the volume of a cone with base radius R and height h, rotated about the ais using the disk method. R h z Take a thin solid disc at arbitrar, with volume V π }{{} area The volume of the cone is V h h h π π π d ( [ R R h ) d R R h + R ] h d π [R R h + 1 R 3 3 h [ π R h R h + 1 ] 3 R h ] h 1 3 πr h 3
Eample: 3.3 Find the volume of a cone with base radius R and height h, rotated about the ais using the shell method. ring z Build up the volume b constructing thin, circular annular shells with radius and thickness The ring area of the shell is given b ring area π and the differential shell volume is V (π ) The total volume of the cone is given as V R πd but can be epressed in terms of as R R h h h R 4
Therefore R V π (h hr ) d π [ 1 1 3 πr h h 1 3 h R 3 ] R 5
Eample: 3.4 Find the surface area of a cone with base radius R and height h, rotated about the ais. R-R h s s z A s Recall that the arc length for () is given as ( ) d ds 1 + d d To find the cone surface area, at an arbitrar, take a slice that has an arc length of s. The surface area for this slice is A S π s The total surface area is given b A s h πds h ( ) d π 1 + d d R R h d d R h 6
Therefore A s h ( π 1 + R ) d h h π (R Rh ) ) 1/ (1 + R d h [ h + R ] 1/ h π (R Rh ) d h [ h + R ] 1/ [ π R 1 ] R h h h π [ h + R ] 1/ [Rh 1 ] Rh h πr R + h } {{ } 1 Rh 7
Eample: 3.5a Find the value for I 4 d using Trapezoidal rule and Simpson s rule We know the eact answer is I eact 5 5 5 5 6.4 Choose n 8 panels of equal width in the range. Therefore h 1/4 Trapezoidal Rule h 1/4 b f() 4 n 8 panels a I h [ ] n 1 f(a) + f(b) + f( i ) 1/4 6.56646 4 + 4 + i1 ( ) 1 4 ( ) 4 ( ) 3 4 ( ) 4 4 ( ) 5 4 ( ) 6 4 ( ) 7 4 + + + + + + 4 4 4 4 4 4 4 The error is ɛ I trap I eact 6.56646 6.4 +.16646 Simpson s Rule I h 3 f(a) + f(b) + 4 n 1 i1,3,5,... f( i ) + n i,4,6,... f( i ) 8
1/4 4 4 + 4 + 4 6.414 ( ) 1 4 ( ) 3 4 ( ) 5 4 ( ) 7 4 ( ) 4 ( ) 4 4 ( ) 6 4 + + + + + + 4 4 4 4 4 4 4 The error is ɛ I simp I eact 6.414 6.4 +.14 Simpson s rule is clearl better for the same value of h. This is usuall the case. 9
Eample: 3.5b Find the value for I 4 d using Romberg integration From the trap rule h n I 1/ 4 I 1 7.65 1/4 8 I 6.5664 1/8 16 I 3 6.4417 h h h 4 h 6 1/ 7.65 4(6.5664) 7.65 6.41 3 16(6.41) 6.41 1/4 6.5664 15 4(6.4417) 6.5664 6.41 3 1/8 6.4417 6.4 where nd column is the result from the Trapezoidal rule with ɛ h 3rd column is the first etrapolation of the elements with ɛ h, this gives an error of ɛ h 4, since we have eliminated the h terms the last column is the second etrapolation using the ɛ h 4 terms to give a result with and error where ɛ h 6 This leads to a ver accurate answer. Can keep going to the right to give (h 8 ) 64( ) ( ) 63 (h 1 ) 56( ) ( ) 55 6 1 8 1 1
Eample: 3.6 Find the area in the + ve quadrant bounded b circles 1 area A to be found - -1 1 + 1 ( 1) + 1 Integration in an coordinate plane would be difficult here due to the shape. r θ is much easier. 1st transform the equations of the circles over to Polar coordinates + 1 r (cos θ + sin θ) 1 }{{} 1 where r cos θ r cos θ r sin θ r sin θ ( 1) + 1 + 1 + 1 r} cos {{ θ} r } cos {{ θ} where +1 + r sin θ }{{} 1 r cos θ r cos θ r cos θ r sin θ r sin θ 11
r (cos θ + sin θ) r cos θ }{{} 1 r1 P r cos r r cos θ r1 r cos θ 3 1 The intersection point is at r 1. Therefore at r 1 find θ. Set r cos θ. Therefore at point P, r 1 and θ π/3. cos θ 1 cos θ 1 θ 6 π 3 radians The bounds of integration are therefore θ π/3 and 1 r cos θ Start with da r dr dθ at arbitrar (r, θ) Build up a wedge slice ( ) cos θ rdr dθ r1 θ fied add up wedges A ( π/3 cos θ θ r1 rdr ) dθ π/3 1 cos θ r 1 dθ 1 π/3 (4 cos θ 1)dθ π/3 A 1 (4 cos θ 1)dθ 4 [ 1 θ + 1 ] π/3 sin θ 1 π/3 4 θ 1
π 3 + 1 sin π 3 π 6 π 6 + 1 sin ( π 3 ).957 where from the integral tables (Schaums 17.18.9) cos θ θ + 1 sin θ 4 As a rough check - 1 4 πr π 4.79 OK 13
Eample: 3.7 Find the surface area in the + ve octant for z f(, ) 4. 4 plane 4 -/ Let F z f(, ) z (4 ) where the limits of integration are : 4 : / The partial derivatives are f 1 f ds 1 + 1 + 4dd 6dd The surface strip area is given b ( ) / 6 d d 14
The surface area is then S 4 / 6 dd 6 R Therefore S 4 6 which is approimatel equal to 9.8 m. 15
Eample: 3.8 Given the sphere, + + z a, derive the formula for surface area. Let F z f(, ) z (a ) 1/ }{{} f(,) z ds + + z a Consider the 1/8 sphere in the + ve octant. The partial derivatives are f 1( ) a f 1( ) a da region + a and the bounds of integration are : (a ) 1/ : a Let the area in the plane (R ) be da dd and the surface above be ds. ds 1 + a + a dd a + + dd a 1 a a dd While we can sum in either direction, we will sum over first The strip surface area is a 1 a a d d 16
with fied. Then we sum up the strips over S a a 1 a a d d Solving the inner integral first, where we let c a. (Schaum s 18.) c S a 1 ( ) c d a c sin 1 c a [ sin 1 (1) sin 1 () ] [ π ] a Therefore the surface area is S a ( a π ) d a π a πa 1/8 of a sphere The total sphere area is S 8 πa 4πa 17
Eample: 3.9 Find the volume formed in the + ve octant between the coordinate planes and the surface z f(, ) 4 4 plane 4 -/ We will determine the volume of the solid, i.e. the solid under the f(, ) surface. We will identif the solid region as R which is the projection of the surface f(, ) downward onto the, plane. Start with a column at arbitrar (, ) in R, where the volume is zda base area height z f(, ) 4 The volume of the column is then V (4 ) }{{} height }{{} da We then sum over (with fied) to get a slice in direction { } / slice volume (4 )d d fied 18
Sum up the slices over V 4 ( / (4 )d ) d The volume under the surface z f(, ) is then V f(, )dd R The order of integration can be reversed, it depends on which ever is easier. Evaluation of the volume integral gives V 4 [ 4 4 4 ( ) / 4 d 4 ( ) ( ) ( ) ] d )] [8 + (4 + d 4 (4 + 14 ) 4 + 3 1 4 d 16 16 + 64 1 16 3 m3 19
Eample: 3.1a Find the mean value of f() sin in the domain to π. We can think of the average value as f, where the area under the curve is f π. f Therefore, fπ π f()d which means f 1 π π f()d or more specificall f 1 π π sin d cos π + cos π.637
Eample: 3.1b Find the mean value of temperature for T f(, ) 4. T of plate at (,) f(,) 4 plate,, -/ fa R f(, )dd R f A Therefore f 1 A R R f(, )dd A R 1 4 4 m R 16 3 f 1 4 16 3 4 3 (from previous eample) 1
Eample: 3.1c Derive the formula for the volume of revolution. for the following sphere: + + z a. z a surface z f(,) a a region + a Consider the + ve octant, i.e. the 1/8 th of the sphere. In polar coordinates we have, z f(r, θ), where z a r Substituting back into the Cartesian formulation for volume we get V a a a dd This is ver mess to work with and leads us over to polar coordinates. A column at an arbitrar (r, θ) has a base of r dr dθ and a height of z a r. If we then integrate over r to build up a wedge slice with θ fied ( a ) slice volume a r rdr dθ r θ fied Now integrate this over θ to add up the slices π/ ( a ) V a r rdr dθ r
The volume of the sphere is then given b V R f(r, θ) rdrdθ }{{}}{{} z area From Schaums 17.11.9 V π/ ( a r ) a r r dr dθ Evaluation gives (Schaum s 17.11.9) V π/ θ (a r ) 3/ 3 a r dθ π/ θ a3 3 θ a 3 3 }{{} wedge volume π/ πa3 6 dθ The total volume of the sphere is 8 times this value V total 8 πa3 6 4 3 πa3 Aside: let u a r and du rdr. Therefore we can write rdr du/. The limits of integration are r u a and r a u. The integral then becomes u 1/ [ a du 3 1 ] 3 u3/ a 1 3 [ a3 ] 1 3 a3 3
Eample: 3.11 Find the volume of the paraboloid, z + for z 4. Consider onl the + ve octant, i.e. 1/4 of the volume. z 4 z + surface sum projection onto (,) plane + 4 Start with a differential volume dv dddz at an arbitrar (,, z) in dv. Build up a column - sum over z ( 4 z + dz ) dd lower bound is the surface of the paraboloid upper limit is the circle at a radius of 4 Build up a slice, and sum over with fied. 4 4 dzd d z + Sum over 4 4 V dzdd z + 4
Evaluate V 4 (4 ) (4 )dd 4 (4 ) 3/ d 1 3 3 3 4 d (4 ) 3/ d 3 (4 ) 3/ d Using the integral tables or through substitution V 3 3π π The total volume is V 4(π) 8π z We could check this value using a volume of revolution dv π dz πzdz 4 z and V 4 z πzdz π z 4 8π o Area dv dz z dz 5
Eample: 3.1 Find the volume bounded b a clinder, + a and a paraboloid, z z + + a z + a a Cartesian Coordinates Consider the positive octant, start with V d d dz We will sum over z(, fied) to make a column. ( ) + vol dz dd z Now sum the columns over ( fied) to make a slice. ( a ) + vol dz d d z Finall, sum the slices over to get the total volume V a a + dz d d z a a ( + )d d ) a ( + 3 a d 3 6
a ( a + 1 ) ( a ) 3/ d 3 4 ( a ) 3/ + a 8 [ a + a sin 1 ] a a πa4 8 The total for all 4 octants is Total V 4 πa4 8 πa4 Clindrical Coordinates (r, θ, z) Start with dv rdr dθ dz We will sum over z to make a column with (r, θ) fied. ( ) r vol dz rdr dθ z Now sum the columns over r with θ fied to make a wedge slice. ( a ) r vol dz rdrdθ }{{} r z planar area Sum the wedges over θ to get V π a r θ r z dzrdr dθ 7
Finall, evaluate from inner to outer V π a θ r π r 4 a θ 4 r rdr dθ dθ a4 π dθ 4 θ π a4 4 πa4 Note how much easier the integrations are when an ai-smmetric problem is solved using clindrical coordinates. 8
Eample: 3.13 Derive a formula for the volume of a sphere with radius, a + + z a In Cartesian coordinates, we will consider the positive octant where z a As in the previous eample, we will build up from da d d, then sum over z to obtain a column, the sum the columns over to obtain a slice, and finall sum the slices. a a a V 8 dzdd 4 z 3 πa3 Note: although not shown, the integrations are mess. Using spherical coordinates, (R, θ, φ). The sphere surface is just R a. Start with dv R sin φ dr dθ dφ Sum over R to make a column, ( R a). Sum the columns over φ to make a wedge ( φ π). The sum the wedges over θ ( θ π) to get the total volume V π π a θ φ R R sin φdr dφ dθ a3 π ( cos φ) π φ 3 dθ θ a3 π dθ 4 3 θ 3 πa3 9
Eample: 3.14 Find the centroid, center of mass and the 1st moment of mass for a quarter circle of radius a with an inner circle of radius a/ made of lead with a densit of ρ 1 11, kg/m 3 and an outer circle of radius a made aluminum with a densit of ρ, 5 kg/m 3. The thickness is uniform throughout at t 1 mm. a + a a/ ρ 1 11 g/cm 11 kg/m 1 ρ.5 g/cm 5 kg/m a/ a Centroid: The total area is given b A R dd 1 4 πa The 1st moment of area about the ais is F R da In Cartesian coordinates a - da d d F a a d d a a d (a ) 3/ 3 a a3 3 If we use polar coordinates with r cos θ and da rdrdθ 3
a - da r dr d F π/ a θ r (r cos θrdr)dθ a3 π/ cos θdθ 3 θ a3 3 sin θ π/ a3 3 Therefore F A a3 /3 (πa )/4 4a 3π.44a F A R dd A 4a 3π.44a (b smmetr) Center of Mass If the thickness, t and the densit, ρ are constant, then the center of mass is equivalent to the centroid. But in this eample we have more mass concentrated toward the origin, so we can epect the center of mass, c, c <,. The total mass is M R ρ(, )tda ρ 1 t da + ρ t R 1 da R [ 1 ( a ) ] ρ 1 t 4 π + ρ t 1 [ ( a ) ] 4 π a ρ 1 tπ a 16 + ρ tπ a 4 ρ tπ a 16 ) πta ( ρ1 ρ 16 πta (ρ 1 + 3ρ ) 16 + ρ 4 31
The 1st moment of mass about the ais dm R ρtda + R 1 ρtda R Clearl R 1 and R are easiest to define in polar coordinates ρ 1 t π/ a/ θ ρ 1 t r π/ ρ 1ta 3 4 θ r cos θrdrdθ + ρ t π/ a θ ra/ r cos θrdrdθ ( 1 a ) [ 3 π/ ( 1 a ) ] 3 cos θdθ + ρ t a 3 cos θdθ 3 θ 3 π/ cos θdθ } {{ } 1 + ρ ta 3 3 ( 1 1 ) π/ cos θdθ 8 }{{} 1 ρ 1ta 3 4 + 7 4 ρ ta 3 ta3 (ρ 1 + 7ρ ) 4 1st moment of mass Therefore c R dm R dm a ( ρ 1 + 7ρ 3π ρ 1 + 3ρ ta 3 4 (ρ 1 + 7ρ ) πta 16 (ρ 1 + 3ρ ) ) a 3π ( ) 11 + 7 5.37a 11 + 3 5 B smmetr c c center of mass.37a a a/ centroid center of center of area.44a mass a/ a 3
Eample: 3.15 Find the area of the paraboloid z + below the plane z 1 The surface area S projects into the interior of the circle + 1 in the plane. This region is denoted in the section above as G. Here z f(, ) + so that f, f and S + 1 4 + 4 + 1d d The double integral above is easier to integrate in polar coordinates, since the combination of + ma be replaced b r. Taking the element area to be da r dr dθ in place of d d, we obtain S r 1 π 1 4r + 1 r dr dθ 4r + 1 r dr dθ π 6 (5 5 1) 33
Eample: 3.16 Find the moment of inertia about the ais of the area enclosed b the cardioid r a(1 cos θ) It is customar to take the densit as unit when working with a geometrical area. Thus we have with I A da r cos θ, da rdrdθ If we integrate first with respect to r, then for an θ between and π, r ma var from to a(1 cos θ) I π a(1 cos θ) π a 4 r 3 cos θdrdθ 4 cos θ(1 cos θ) 4 dθ The integral above can be solved using the reduction formula π cos n θdθ cosn 1 θ sin θ n π + n 1 π cos n θdθ n to give I a4 π (cos θ 4 cos 3 θ + 6 cos 4 θ 4 cos 5 θ + cos 6 θ) dθ 4 [ ] a4 4 49πa4 3 1 + 18 4 + 5 8 π 34
Eample: 3.17 Find the center of gravit of a homogeneous solid hemisphere of radius a We ma choose the origin at the center of a sphere and consider the hemisphere that lies above the plane. The equation of the hemispherical surface is z a or in terms of clindrical coordinates z a r B smmetr we have We calculate z: z zdv dv π a a r 3 πa3 3a 8 35
Eample 4.1 Show for the 3D case f(,, z) that curl grad f f î f + ĵ f + ˆk f z holds for an function f(,, z), for instance f(,, z) + + sin + z f î f + ĵ f + ˆk f z î( + cos ) + ĵ( + sin ) + ˆk(z) ( f) î ĵ ˆk / / / z + cos + sin z î( ) ĵ( ) + ˆk(cos cos ) 1
Eample 4. Given a 3D temperature field T f(,, z) 8 + 6 + 3z find the average temperature, T along a line from (,, ) to (1, 1, 1). z curve C Parametrize so t increases in the direction of travel. C t, t, z t, t 1 The length of C is given b ) 1 ( d ( ) d ( ) dz L + + dt 3 t dt dt dt The line integral is given b C fds 1 t (8t + 6t + 3t) }{{}} 1 + {{ 1 + 1dt } f value on C ds on C 1 3 (38t + 6t )dt 3 38 t + 1 6t3 1 3 3 The average temperature is T 1 3 3 1 C
Eample: 4.3a Suppose the temperature near the floor of a room (sa at z 1) is described b T f(, ) + 3 where 5 5 4 4 What is the average temperature along the straight line path from A(, ) to B(4, 3). T C f(, )ds ds C where C is the curve (t) t (t) 3 4 t for t 4, where t increases in the direction of travel. The numerator is C f(, )ds 4 t 4 t ) ( d ( ) d f[(t), (t)] + dt dt dt (t + 916 ) t 1 + 9 dt 86.1 3 16 The denominator is the arc length L ) 4 ( d ( ) d L + dt 5 t dt dt The mean temperature is T 86.1 5 17. 3
Eample: 4.3b What is the average room temperature along the walls of the room? T C f(, )ds ds C where the closed curve C is defined in 4 sections C 1 4 t 5 t 5 C 5 t 4 t 4 C 3 4 5 t t 1 C 4 5 8 t t 8 Note: Set up so that t increases in the direction of travel. fds C 1 fds + C 1 fds + C fds + C 3 fds C 4 C 1 fds 5 t 5 { t + 16 3 } 1 + dt 118.9 Similarl C 8.7 C 3 118.9 C 4 8.7 The room perimeter, ds, in this instance is 36. The mean temperature is given b T 118.9 + 8.7 + 118.9 + 8.7 36 11. 4
Eample: 4.3c What is the average temperature around a closed circular path + 9? T C f(, )ds ds where C is a closed circular path C (t) 3 cos t (t) 3 sin t for t π. Show that this integration gives T 9 3 17 as it should for this f(, ), i.e. curve C is actuall the T 17 contour. r cos θ r sin θ let θ t 5
Eample 4.4 Given a force field in 3D: F î(3 6) + ĵ( + 3z) + ˆk(1 4z ) What is the work done b F on a particle (i.e. energ added to the particle) if it moves in a straight line from (,, ) to (1, 1, 1) through the force field. W F d r P d + Qd + Rdz C (3 6z)d + ( + 3z)d + (1 4z )dz C the parametric form of C leads to: t t z t d dt d dt dz dt for t 1. Therefore W 1 t (3t 6t )dt + (t + 3t )dt + (1 4t 4 )dt 1 t (1 + t 4t 4 )dt (t + t 4 5 t ) 1 + 1 4 5 6 5 Joules 1 6
Eample: 4.5a The gravitational force on a mass, m, due to mass, M, at the origin is F G Mm r r 3 K r r 3 where K GMm. The vector field is given b: where F (,, z) îp + ĵq + ˆkR K P ( + + z ) 3/ K Q ( + + z ) 3/ Kz R ( + + z ) 3/ Compute the work, W, if the mass, m moves from A to B along a semi-circular path in the (, z) plane: + z 16 or z From A(,.1, 1.61) to B(, 16, ) 16 and z m curve C in the plane (z plane) - circle, radius 8 center (,8,) A M B W C F d r C (îp + ĵq + ˆkR) (îd + ĵd + ˆkd) The integrand can either be rewritten using the parametric equations for curve, C, or can be rewritten in terms of the variable here since we have an eplicit equation for C. 7
On curve C P K ( + + z ) 3/ Q K ( + + z ) K 3/ ( + 16 ) K 3/ 64 R Kz 16 K ( + + z ) 3/ 64 3/ Also d dz 1 (16 )d 16 Therefore W C 16 P d + Qd + Rdz.1 16.1 K 64 d K 16 16 64 3/ 16 d K ( 64 1 + 8 ) d K 4 1 16.1.78K Joules Energ of.78k Joules must be supplied to move mass m along this line from A to B i.e to counteract gravit. 8
Eample: 4.5b Find the work to move through the same field, but following a straight line path from A(,.1, 1.61) to B(, 16, ). z m A curve C M B Compute W, epressing the path C in parametric form. The equation of C is: z 1.61 1.61 (.1) and 16.1 or in parametric form t z 1.69.793t for.1 t 16 W C P d + Qd + Rdz K P d ( + + z ) 3/ K Q ( + + z ) Kt d dt 3/ {t + (1.69.793t) } 3/ Kz R ( + + z ) K(1.69.793t) dz.793dt 3/ {t + (1.69.793t) } 3/ Substitution and simplification ields 16 W K t.1.78k J oules.11.994t (1.63t.1t + 1.61) 3/ dt This is an eample of a conservative force field - work W is the same for all paths between A and B. 9
Eample: 4.6 Suppose the temperature variation (same for all (, )) in the atmosphere near the ground is T (z) 4 z 5 where T is in C and z is in m. Look at a clindrical building roof as follows: z 1 m high 3 m long 1 m What is the air temperature in contact with the roof? The equation of the roof is + z 1 3 The temperature equation in space is T f(,, z) 4 z and the roof equation for S is z g(, ) 5 1 3 The average temperature over the roof is T S f(,, z)ds Area of S 1
z zg(,) 1-3 d d 1 projection is rectangular here Consider half of the roof - i.e. the positive octant S fds where f 4 z 5 4 1 5 (1 ) + 1 5 ds ( ) g ( ) g 1 + + dd 1 + + 1 1 1 11
Therefore S fds 1 3 ( + 1 5 ) 1 1 dd 1 1 3 ( + ) d 1 We can either go to the tables or use numerical integration. For instance we could use Simpson s rule to obtain 1 () d 465.6 The surface area is S ds 1 4 clinder 1 4 πrh π (1)(3) 471. T 3(465.6) 471. 9.6 C 1
Eample: 4.7 Given a velocit field in 3D space find V î( + z) + ĵ( ) + ˆk(z) a) the flow rate Q (m 3 /s) across the surface z 1 for 1 and 1 in the + ve z direction b) the average velocit across the surface z 1 ds surface S 1 1 Given surface S, with z g(, ) 1 or G z 1, and ds being a surface element and ˆn a unit normal vector given as ˆn ± G G î G + ĵ G G + ˆk z ˆk(1) Choose ˆn ˆk for flow in the + ve direction. Q S V ˆn ds where ( ) g ( ) g ds 1 + + dd dd 13
and V ˆn V ˆk z On surface z 1 V î( + 1) + ĵ( ) + ˆk n V The V changes in magnitude and direction over S. Take V ˆn components for each ds and add up to get total flow. Q V ˆn ds S dd R 1 1 1 dd 1 1 m3 /s b) average velocit across S V avg Q across S Area of S 1 m/s 1 m 1 m/s 14
Eample: 4.8 Given V î(1 + ) + ĵ(1 + ) + ˆk(1 + z 3 ) verif the divergence theorem for a cube, where,, z 1 V ˆndS ( V )dv S V i.e. show that where S cube surface (closed) V interior volume of the cube For a general surface z g(, ) we can use a projection of this surface to obtain ds ( ) g ( ) g ds 1 + + dd Since all surface are just z const. we get g g Which gives that ds dd. Similarl there is no need for ˆn G G contributions. 1. at the face: ˆn(outward) ĵ ds ddz V ˆn (1 + ) 1 since S V ˆndS 1 1 z 1dzd 1 This means that 1 m 3 /s inflow i.e. - ve direction 15
. at the 1 face: ˆn(outward) ĵ ds ddz V ˆn +(1 + ) since 1 S V ˆndS 1 1 z dzd This means that m 3 /s outflow i.e. + ve direction 3. at the z face: ˆn(outward) ˆk ds dd V ˆn (1 + z 3 ) 1 S V ˆndS 1 1 1dd 1 4. at the z 1 face: ˆn(outward) ˆk ds dd V ˆn +(1 + z 3 ) since 1 S V ˆndS 1 1 dd 5. at the face: ˆn(outward) î ds ddz V ˆn (1 + ) 1 since S V ˆndS 1 1 z 1dzd 1 This means that 1 m 3 /s inflow i.e. - ve direction 6. at the 1 face: ˆn(outward) î ds ddz V ˆn +(1 + ) since 1 S V ˆndS 1 1 z dzd This means that m 3 /s outflow i.e. + ve direction 16
Summing up these components, we get S V nds 1 + 1 + 1 + 3 m 3 /s The right hand side is given as V ( V )dv where V u + v + w z 1 + + 3z The RHS integral is solved as 1 1 1 z (1 + + 3z )dzdd 1 1 1 1 1 1 ( z + z + z 3 ) 1 ( + ) dd ( + ) 1 d dd 3 d 3 1 3 m3 /s 17
Eample: 4.8 Given: F î + ĵz + ˆk (a force field in 3D). The closed path C is given b the intersection of: + 4 z 4 The object moves once in a CW direction around C starting at (,, ). Verif Stoke theorem: F d r ( F ) ˆn ds C S Aside: Check the curl of F F î ĵ ˆk z z î(1 ) + ĵ() + ˆk() î We see that we cannot make the following assumption F d r It ma or ma not be dependent of C. The LHS is given as P d + Qd + Rdz C 18
(±) cos t + cos t (±) sin t sin t z 4 cos t + sin t d sin tdt d cos tdt dz sin tdt + cos tdt Note the sign convention due to the CW motion. P cos t on C Q z 8 4 cos t + 4 sin t on C R sin t on C π t ( 4 sin t cos t 16 cos t + 8 cos t 8 sin t cos t 4 sin t 4 sin t cos t)dt π t 16 sin t ( 16 sin t cos t 16 cos t 4 sin t + 8 cos t)dt π ( t 16 sin t π sin t 4 4 ) π ( t + 8 + sin t 4 ) π 4 π + 8 π 4π + 8π 4π The RHS is S ( F ) nds We can choose an S with C as the boundar. Here we will select z g(, ) 4 G z 4 + + 19
F î ĵ ˆk z z î(1 ) + ĵ() + ˆk() î ˆn (±) G G (±) +î + ĵ + ˆk 1 + 1 + 1 î 3 ĵ 3 ˆk 3 Therefore ( F ) ˆn + 1 3 ds ( ) g ( ) g 1 + + dd 3 dd Therefore S R ( R rdrdθ + 1 3 ) 3dd R dd } {{ } area of R r r π 4 π 4π θ θ π rdrdθ LHS RHS, both are positive. Work is done b the F field on the object.
Eample: 4.9 Given a D force field, F (, ) î( 3 ) + ĵ( ) and a path C in the field: Verif Green s theorem with C P d + Qd P 3 Q R ( Q P ) dd LHS: For the general curve C, we need the parametric form to compute C P d+qd. However, because of the shape of C in this eample, a parametric representation of the curve is not required. 1. on C 1 : 1 d P Q W P d + Qd C 1. on C : 1 d 1 1 d P Q P d + Qd C 3. on C 3 : 1 d P Q 1 P d + Qd d 1 1 C 3 4. on C 4 : 1 d P 3 Q 1 P d + Qd d 1 C 4 1 1 1
Therefore W C P d + Qd 1 1 1 Joules A negative value of W implies that work has to be supplied b the object. RHS: R ( Q P ) ( dd ) 3 dd R In this particular case the calculation of the double integral is made easier because R limits are all constants. 1 [ 1 1 ] ( 3 )d d 1 [ ( 3 ) ] 1 d 1 1 ( )d 1 1 Joules Therefore LHS RHS. We can use Green s theorem to change a line integral computation of W to a R instead.