Članki

15.4: Aplikacije dvojnih integralov


Učni cilji

  • Prepozna, kdaj je funkcija dveh spremenljivk integrirana v pravokotni regiji.
  • Prepoznajte in uporabite nekatere lastnosti dvojnih integralov.
  • Ocenite dvojni integral nad pravokotnim območjem, tako da ga zapišete kot ponovljeni integral.
  • Z dvojnim integralom izračunajte površino območja, prostornino pod površino ali povprečno vrednost funkcije nad ravnino.

V tem poglavju preučujemo dvojne integrale in prikazujemo, kako jih lahko uporabimo za iskanje prostornine trdne snovi v pravokotni regiji v ravnini xy. Številne lastnosti dvojnih integralov so podobne tistim, ki smo jih že obravnavali pri enojnih integralih.

Volumni in dvojni integrali

Začnemo z obravnavo prostora nad pravokotnim območjem (R ). Razmislite o neprekinjeni funkciji (f (x, y) ≥0 ) dveh spremenljivk, definiranih na zaprtem pravokotniku (R ):

[R = [a, b] krat [c, d] = levo {(x, y) ∈ mathbb {R} ^ 2 | , a ≤ x ≤ b, , c ≤ y ≤ d desno } nešteto ]

Tu ([a, b] krat [c, d] ) označuje kartezični zmnožek dveh zaprtih intervalov ([a, b] ) in ([c, d] ). Sestavljen je iz pravokotnih parov ((x, y) ), ki sta (a≤x≤b ) in (c≤y≤d ). Graf (f ) predstavlja površino nad ravnino (xy ) - z enačbo (z = f (x, y) ), kjer je (z ) višina površine v točki ((x, y) ). Naj bo (S ) trdna snov, ki leži nad (R ) in pod grafom (f ) (slika ( PageIndex {1} )). Osnova trdne snovi je pravokotnik (R ) v ravnini (xy ). Poiskati želimo prostornino (V ) trdne snovi (S ).

Območje (R ) razdelimo na majhne pravokotnike (R_ {ij} ), vsak s površino (ΔA ) in s stranicama (Δx ) in (Δy ) (slika ( PageIndex { 2} )). To naredimo tako, da razdelimo interval ([a, b] ) na podintervale (m ) in razdelimo interval ([c, d] ) na podintervale (n ). Od tod ( Delta x = frac {b - a} {m} ), ( Delta y = frac {d - c} {n} ) in ( Delta A = Delta x Delta y ).

Prostornina tankega pravokotnega polja nad (R_ {ij} ) je (f (x_ {ij} ^ *, , y_ {ij} ^ *) , Delta A ), kjer ( (x_ {ij} ^ *, , y_ {ij} ^ * )) je poljubna vzorčna točka v vsakem (R_ {ij} ), kot je prikazano na naslednji sliki, (f (x_ {ij} ^ *, , y_ {ij} ^ *) ) je višina ustreznega tankega pravokotnega polja, ( Delta A ) pa površina vsakega pravokotnika (R_ {ij} ).

Z uporabo iste ideje za vse podrekotnike dobimo približno prostornino trdne snovi S kot

[V približno vsota_ {i = 1} ^ m vsota_ {j = 1} ^ n f (x_ {ij} ^ *, , y_ {ij} ^ *) Delta A. nešteto ]

Ta vsota je znana kot dvojna Riemannova vsota in se lahko uporablja za približanje vrednosti prostornine trdne snovi. Tu dvojna vsota pomeni, da za vsak podpravokotnik ocenimo funkcijo na izbrani točki, pomnožimo s površino vsakega pravokotnika in nato dodamo vse rezultate.

Kot smo videli v primeru z eno spremenljivko, dobimo boljši približek dejanski prostornini, če (m ) in (n ) postaneta večji.

[V = lim_ {m, n rightarrow infty} sum_ {i = 1} ^ m sum_ {j = 1} ^ nf (x_ {ij} ^ *, , y_ {ij} ^ *) Delta A neštevilno ]

ali

[V = lim _ { Delta x, , Delta y rightarrow 0} sum_ {i = 1} ^ m sum_ {j = 1} ^ nf (x_ {ij} ^ *, , y_ { ij} ^ *) Delta A. neštevilno ]

Upoštevajte, da se vsota v obeh primerih približuje meji, meja pa je prostornina trdne snovi z osnovo (R ). Zdaj smo pripravljeni definirati dvojni integral.

Definicija

Dvojni integral funkcije (f (x, , y) ) nad pravokotnim območjem (R ) v ravnini (xy ) je opredeljen kot

[ iint_R f (x, , y) dA = lim_ {m, n rightarrow infty} sum_ {i = 1} ^ m sum_ {j = 1} ^ nf (x_ {ij} ^ * , , y_ {ij} ^ *) Delta A. ]

Če je (f (x, y) geq 0 ), potem je prostornina (V ) trdne snovi (S ), ki leži nad (R ) v ravnini (xy ) - in pod grafom (f ) je dvojni integral funkcije (f (x, y) ) nad pravokotnikom (R ). Če je funkcija kdaj negativna, lahko dvojni integral štejemo za "podpisan" volumen na podoben način, kot smo definirali neto podpisano območje v Določenem integralu.

Primer ( PageIndex {1} ): Nastavitev dvojnega integrala in njegovo približevanje z dvojnimi vsotami

Razmislite o funkciji (z = f (x, , y) = 3x ^ 2 - y ) nad pravokotnim območjem (R = [0, 2] krat [0, 2] ) (Slika ( PageIndex {4} )).

  1. Nastavite dvojni integral za iskanje vrednosti podpisanega volumna trdne snovi (S ), ki leži nad (R ) in "pod" grafom (f ).
  2. (R ) razdelite na štiri kvadratke z (m = n = 2 ) in izberite vzorčno točko kot zgornjo desno kotno točko vsakega kvadrata (1,1), (2,1), (1,2 ) in (2,2) (slika ( PageIndex {4} )) za približek podpisane prostornine trdne snovi (S ), ki leži nad (R ) in "pod" grafom ( f ).
  3. (R ) razdelite na štiri kvadratke z (m = n = 2 ) in izberite vzorčno točko kot sredino vsakega kvadrata: (1/2, 1/2), (3/2, 1/2 ), (1 / 2,3 / 2) in (3/2, 3/2) za približevanje podpisanega obsega.

Rešitev

  1. Kot lahko vidimo, je funkcija (z = f (x, y) = 3x ^ 2 - y ) nad ravnino. Da bi našli podpisani volumen (S ), moramo regijo (R ) razdeliti na majhne pravokotnike (R_ {ij} ), vsak s površino (ΔA ) in s stranicami (Δx ) in (Δy ) ter izberite ((x_ {ij} ^ *, y_ {ij} ^ *) ) kot vzorčni točki v vsaki (R_ {ij} ). Tako je dvojni integral nastavljen kot

    [V = iint_R (3x ^ 2 - y) dA = lim_ {m, n → ∞} sum_ {i = 1} ^ m sum_ {j = 1} ^ n [3 (x_ {ij} ^ *) ^ 2 - y_ {ij} ^ *] Delta A. neštevilno ]

  2. Približanje podpisanega volumna z uporabo Riemannove vsote z (m = n = 2 ) imamo ( Delta A = Delta x Delta y = 1 krat 1 = 1 ). Vzorčne točke so tudi (1, 1), (2, 1), (1, 2) in (2, 2), kot je prikazano na naslednji sliki.

Torej,

[ začeti {poravnati *} V & približno vsota_ {i = 1} ^ 2 vsota_ {j = 1} ^ 2 f (x_ {ij} ^ *, y_ {ij} ^ *) Delta A [4pt]
& = sum_ {i = 1} ^ 2 (f (x_ {i1} ^ *, y_ {i1} ^ *) + f (x_ {i2} ^ *, y_ {i2} ^ *)) Delta A [4pt]
& = f (x_ {11} ^ *, y_ {11} ^ *) Delta A + f (x_ {21} ^ *, y_ {21} ^ *) Delta A + f (x_ {12} ^ * , y_ {12} ^ *) Delta A + f (x_ {22} ^ *, y_ {22} ^ *) Delta A [4pt]
& = f (1,1) (1) + f (2,1) (1) + f (1,2) (1) + f (2,2) (1) [4pt]
& = (3 - 1) (1) + (12 - 1) (1) + (3 - 2) (1) + (12 - 2) (1) [4pt]
& = 2 + 11 + 1 + 10 = 24. konec {poravnaj *} ]

  1. Približanje podpisanega volumna z uporabo Riemannove vsote z (m = n = 2 ) imamo ( Delta A = Delta x Delta y = 1 krat 1 = 1 ). V tem primeru so vzorčne točke (1/2, 1/2), (3/2, 1/2), (1/2, 3/2) in (3/2, 3/2).
    Torej,
    [ začeti {poravnati *} V & približno vsota_ {i = 1} ^ 2 vsota_ {j = 1} ^ 2 f (x_ {ij} ^ *, y_ {ij} ^ *) Delta A [4pt]
    & = f (x_ {11} ^ *, y_ {11} ^ *) Delta A + f (x_ {21} ^ *, y_ {21} ^ *) Delta A + f (x_ {12} ^ * , y_ {12} ^ *) Delta A + f (x_ {22} ^ *, y_ {22} ^ *) Delta A [4pt]
    & = f (1 / 2,1 / 2) (1) + f (3 / 2,1 / 2) (1) + f (1 / 2,3 / 2) (1) + f (3/2, 3/2) (1) [4pt]
    & = levo ( frac {3} {4} - frac {1} {4} desno) (1) + levo ( frac {27} {4} - frac {1} {2} desno) (1) + levo ( frac {3} {4} - frac {3} {2} desno) (1) + levo ( frac {27} {4} - frac {3} {2} desno) (1) [4pt]
    & = frac {2} {4} + frac {25} {4} + levo (- frac {3} {4} desno) + frac {21} {4} = frac {45} {4} = 11. konec {poravnaj *} ]

Analiza

Upoštevajte, da se približni odgovori razlikujejo zaradi izbire vzorčnih točk. V obeh primerih navajamo nekaj napak, ker uporabljamo le nekaj vzorčnih točk. Tako moramo raziskati, kako lahko dosežemo natančen odgovor.

Vadba ( PageIndex {1} )

Uporabite isto funkcijo (z = f (x, y) = 3x ^ 2 - y ) nad pravokotnim območjem (R = [0,2] × [0,2] ).

(R ) razdelite na iste štiri kvadratke z (m = n = 2 ) in izberite vzorčne točke kot zgornjo levo točko vsakega kvadrata (0,1), (1,1), (0 , 2) in (1,2) (slika ( PageIndex {5} )) za približek podpisane prostornine trdne snovi (S ), ki leži nad (R ) in "pod" grafom (f ).

Namig

Sledite korakom iz prejšnjega primera.

Odgovorite

[V približno vsota_ {i = 1} ^ 2 vsota_ {j = 1} ^ 2 f (x_ {ij} ^ *, y_ {ij} ^ *) , Delta A = 0 nešteto ]

Upoštevajte, da smo koncept dvojnega integrala razvili s pomočjo pravokotnega območja (R ). Ta koncept je mogoče razširiti na katero koli splošno regijo. Kadar pa območje ni pravokotno, se podkotniki morda ne prilegajo popolnoma (R ), zlasti če je osnovno območje ukrivljeno. To situacijo podrobneje preučimo v naslednjem razdelku, kjer preučujemo regije, ki niso vedno pravokotne in podkotniki morda ne ustrezajo popolnoma regiji (R ). Tudi višine morda niso natančne, če je površina (z = f (x, y) ) ukrivljena. Vendar se napake na straneh in višina, pri kateri se kosi morda ne prilegajo popolnoma znotraj trdnega (S ), približujejo 0, saj se (m ) in (n ) približujeta neskončnosti. Obstaja tudi dvojni integral funkcije (z = f (x, y) ), pod pogojem, da funkcija (f ) ni preveč prekinjena. Če je funkcija omejena in neprekinjena nad (R ), razen na končnem številu gladkih krivulj, potem dvojni integral obstaja in pravimo, da je ff integrabilno nad (R ).

Ker ( Delta A = Delta x Delta y = Delta y Delta x ), lahko izrazimo (dA ) kot (dx , dy ) ali (dy , dx ). To pomeni, da kadar uporabljamo pravokotne koordinate, je dvojni integral nad območjem (R ) označen z

[ iint_R f (x, y) , dA neštevilno ]

lahko zapišemo kot

[ iint_R f (x, y) , dx , dy neštevilno ]

ali

[ iint_R f (x, y) , dy , dx. neštevilno ]

Zdaj pa naštejmo nekaj lastnosti, ki so lahko koristne za izračun dvojnih integralov.

Lastnosti dvojnih integralov

Lastnosti dvojnih integralov so zelo koristne pri njihovem računanju ali drugačnem delu z njimi. Tukaj navajamo šest lastnosti dvojnih integralov. Lastnosti 1 in 2 se imenujejo linearnost integrala, lastnost 3 je aditivnost integrala, lastnost 4 je monotonost integrala, lastnost 5 pa se uporablja za iskanje meja integrala. Lastnost 6 se uporablja, če je (f (x, y) ) produkt dveh funkcij (g (x) ) in (h (y) ).

Izrek: LASTNOSTI DVOJNIH INTEGRALOV

Predpostavimo, da sta funkciji (f (x, y) ) in (g (x, y) ) združljivi v pravokotni regiji (R ); (S ) in (T ) sta podregiji (R ); in predpostavimo, da sta (m ) in (M ) realni številki.

  1. Vsota (f (x, y) + g (x, y) ) je integrabilna in

[ iint_R [f (x, y) + g (x, y)] , dA = iint_R f (x, y) , dA + iint_R g (x, y) , dA. neštevilno ]

  1. Če c je konstanta, potem je (cf (x, y) ) integrabilno in

[ iint_R cf (x, y) , dA = c iint_R f (x, y) , dA. neštevilno ]

  1. Če (R = S∪T ) in (S∩T = ∅ ) razen prekrivanja meja, potem

[ iint_R f (x, y) , dA = iint_S f (x, y) , dA + iint_T f (x, y) , dA. neštevilno ]

  1. Če je (f (x, y) geq g (x, y) ) za ((x, y) ) v (R ), potem

[ iint_R f (x, y) , dA geq iint_R g (x, y) , dA. neštevilno ]

  1. Če (m leq f (x, y) leq M ) in (A (R) = , besedilo {območje} , R ), potem

[m cdot A (R) leq iint_R f (x, y) , dA leq M cdot A (R). neštevilno ]

  1. V primeru, da je mogoče (f (x, y) ) razložiti kot zmnožek funkcije (g (x) ) samo (x ) in funkcije (h (y) ) od samo (y ), nato čez regijo (R = velika {(x, y) , | , a leq x leq b, , c leq y leq d velika } ), dvojni integral lahko zapišemo kot

[ iint_R f (x, y) , dA = levo ( int_a ^ b g (x) , dx desno) levo ( int_c ^ d h (y) , dy desno). neštevilno ]

Te lastnosti se uporabljajo pri vrednotenju dvojnih integralov, kot bomo videli kasneje. Ko se bomo seznanili z računskimi orodji dvojnih integralov, bomo postali vešči uporabe teh lastnosti. Pojdimo torej do tega zdaj.

Ponovljeni integrali

Do zdaj smo videli, kako nastaviti dvojni integral in kako zanj pridobiti približno vrednost. Lahko si tudi predstavljamo, da je ocenjevanje dvojnih integralov z uporabo definicije lahko zelo dolgotrajen postopek, če izberemo večje vrednosti za (m ) in (n ), zato potrebujemo praktično in priročno tehniko za izračun dvojnih integralov. Z drugimi besedami, naučiti se moramo, kako izračunati dvojne integrale, ne da bi uporabili definicijo, ki uporablja meje in dvojne vsote.

Osnovna ideja je, da je vrednotenje lažje, če lahko dvojni integral razbijemo v enojne integrale tako, da najprej integriramo glede na eno spremenljivko in nato še na drugo. Ključno orodje, ki ga potrebujemo, se imenuje ponovljeni integral.

Definicije: ponovljeni integrali

Predpostavimo, da so (a ), (b ), (c ) in (d ) realna števila. Določimo ponovljeni integral za funkcijo (f (x, y) ) nad pravokotnim območjem (R = [a, b] × [c, d] ) kot

[ int_a ^ b int_c ^ d f (x, y) , dy , dx = int_a ^ b levo [ int_c ^ d f (x, y) , dy desno] dx ]

ali

[ int_c ^ d int_a ^ b f (x, y) , dx , dy = int_c ^ d levo [ int_a ^ b f (x, y) , dx desno] dy. ]

Zapis ( int_a ^ b left [ int_c ^ df (x, y) , dy right] dx ) pomeni, da (f (x, y) ) integriramo glede na (y ), medtem ko (x ) držimo konstantno. Podobno zapis ( int_c ^ d left [ int_a ^ bf (x, y) , dx right] dy ) pomeni, da integriramo (f (x, y) ) glede na ( x ), medtem ko (y ) držite konstantno. Dejstvo, da je mogoče dvojne integrale razdeliti na ponovljene integrale, je izraženo v Fubinijevem izreku. Ta izrek si predstavljajte kot bistveno orodje za ocenjevanje dvojnih integralov.

Izrek: FUBINIJEV TEOREM

Denimo, da je (f (x, y) ) funkcija dveh spremenljivk, ki je neprekinjena nad pravokotnim območjem (R = big {(x, y) ∈ mathbb {R} ^ 2 | , a leq x leq b, , c leq y leq d velik } ). Nato na sliki ( PageIndex {6} ) vidimo, da je dvojni integral (f ) nad regijo enak ponovljenemu integralu,

[ iint_R f (x, y) , dA = iint_R f (x, y) , dx , dy = int_a ^ b int_c ^ df (x, y) , dy , dx = int_c ^ d int_a ^ bf (x, y) , dx , dy. ]

Na splošno velja, da Fubinijev izrek velja, če je (f ) omejen na (R ) in (f ) prekinjen le na končnem številu zveznih krivulj. Z drugimi besedami, (f ) mora biti integriran nad (R ).

Primer ( PageIndex {2} ): Uporaba Fubinijevega izrek

Uporabite Fubinijev izrek za izračun dvojnega integrala ( displaystyle iint_R f (x, y) , dA ), kjer (f (x, y) = x ) in (R = [0, 2] krat [0, 1] ).

Rešitev

Fubinijev izrek ponuja enostavnejši način vrednotenja dvojnega integrala z uporabo ponovljenega integrala. Upoštevajte, kako mejne vrednosti regije (R ) postanejo zgornja in spodnja meja integracije.

[ začeti {poravnati *} iint_R f (x, y) , dA & = iint_R f (x, y) , dx , dy [4pt]
& = int_ {y = 0} ^ {y = 1} int_ {x = 0} ^ {x = 2} x , dx , dy [4pt]
& = int_ {y = 0} ^ {y = 1} levo [ frac {x ^ 2} {2} bigg | _ {x = 0} ^ {x = 2} desno] , dy [4pt]
& = int_ {y = 0} ^ {y = 1} 2 , dy = 2y bigg | _ {y = 0} ^ {y = 1} = 2 end {align *} ]

Dvojna integracija v tem primeru je dovolj preprosta, da neposredno uporabimo Fubinijev izrek, kar nam omogoča pretvorbo dvojnega integrala v ponovljeni integral. Posledično smo zdaj pripravljeni pretvoriti vse dvojne integrale v ponovljene integrale in pokazati, kako nam lahko prej naštete lastnosti pomagajo oceniti dvojne integrale, kadar je funkcija (f (x, y) ) bolj zapletena. Upoštevajte, da je vrstni red integracije mogoče spremeniti (glejte primer 7).

Primer ( PageIndex {3} ): ponazoritev lastnosti i in ii

Ocenite dvojni integral [ iint_R (xy - 3xy ^ 2) , dA, , text {kjer} , R = velik {(x, y) , | , 0 leq x leq 2, , 1 leq y leq 2 velika }. Neštevilna ]

Rešitev

Ta funkcija ima dva dela: en del je (xy ), drugi pa (3xy ^ 2 ). Tudi drugi del ima konstanto 3. Upoštevajte, kako uporabljamo lastnosti i in ii za lažjo oceno dvojnega integrala.

[ začeti {poravnati *} iint_R (xy - 3xy ^ 2) , dA & = iint_R xy , dA + iint_R (-3xy ^ 2) , dA & & text {Lastnost i: Integral od vsota je vsota integralov.} [4pt]
& = int_ {y = 1} ^ {y = 2} int_ {x = 0} ^ {x = 2} xy , dx , dy - int_ {y = 1} ^ {y = 2} int_ {x = 0} ^ {x = 2} 3xy ^ 2 , dx , dy & & text {Pretvorite dvojne integrale v ponovljene integrale.} [4pt]
& = int_ {y = 1} ^ {y = 2} levo ( frac {x ^ 2} {2} y desno) bigg | _ {x = 0} ^ {x = 2} , dy - 3 int_ {y = 1} ^ {y = 2} levo ( frac {x ^ 2} {2} y ^ 2 desno) bigg | _ {x = 0} ^ {x = 2} , dy & & text {Integriraj glede na $ x $, pri čemer ima $ y $ konstanto.} [4pt]
& = int_ {y = 1} ^ {y = 2} 2y , dy - int_ {y = 1} ^ {y = 2} 6y ^ 2 dy & & text {Lastnost ii: umestitev konstante pred integral.} [4pt]
& = 2 int_1 ^ 2 y , dy - 6 int_1 ^ 2 y ^ 2 , dy & & text {Integriraj glede na y.} [4pt]
& = 2 frac {y ^ 2} {2} bigg | _1 ^ 2 - 6 frac {y ^ 3} {3} bigg | _1 ^ 2 [4pt]
& = y ^ 2 bigg | _1 ^ 2 - 2y ^ 3 bigg | _1 ^ 2 [4pt]
& = (4−1) - 2 (8−1) = 3 - 2 (7) = 3 - 14 = −11. end {poravnaj *} ]

Primer ( PageIndex {4} ): Ilustriranje lastnosti v.

Nad regijo (R = velika {(x, y) , | , 1 leq x leq 3, , 1 leq y leq 2 velika } ) imamo (2 leq x ^ 2 + y ^ 2 leq 13 ). Poiščite spodnjo in zgornjo mejo za integral ( displaystyle iint_R (x ^ 2 + y ^ 2) , dA. )

Rešitev

Za spodnjo mejo integrirajte konstantno funkcijo 2 po območju (R ). Za zgornjo mejo integrirajte konstantno funkcijo 13 po območju (R ).

[ začeti {poravnati *} int_1 ^ 2 int_1 ^ 3 2 , dx , dy & = int_1 ^ 2 [2x bigg | _1 ^ 3] , dy = int_1 ^ 2 2 (2) dy = 4y bigg | _1 ^ 2 = 4 (2 - 1) = 4 [4pt] int_1 ^ 2 int_1 ^ 3 13dx , dy & = int_1 ^ 2 [13x bigg | _1 ^ 3] , dy = int_1 ^ 2 13 (2) , dy = 26y bigg | _1 ^ 2 = 26 (2 - 1) = 26. konec {poravnaj *} ]

Tako dobimo ( displaystyle 4 leq iint_R (x ^ 2 + y ^ 2) , dA leq 26. )

Primer ( PageIndex {5} ): ponazoritev lastnosti vi

Izračunajte integral ( displaystyle iint_R e ^ y cos x , dA ) nad območjem (R = velik {(x, y) , | , 0 leq x leq frac { pi} {2}, , 0 leq y leq 1 velika } ).

Rešitev

To je odličen primer za lastnost vi, ker je funkcija (f (x, y) ) očitno produkt dveh funkcij z eno spremenljivko (e ^ y ) in ( cos x ). Tako lahko integral razdelimo na dva dela in nato vsakega integriramo kot problem integracije z eno spremenljivko.

[ začeti {poravnati *} iint_R e ^ y cos x , dA & = int_0 ^ 1 int_0 ^ { pi / 2} e ^ y cos x , dx , dy [4pt ]
& = levo ( int_0 ^ 1 e ^ y dy desno) levo ( int_0 ^ { pi / 2} cos x , dx desno) [4pt]
& = (e ^ y bigg | _0 ^ 1) ( sin x bigg | _0 ^ { pi / 2}) [4pt]
& = e - 1. konec {poravnaj *} ]

Vadba ( PageIndex {2} )

a. Za vrednotenje integrala uporabite lastnosti dvojnega integrala in Fubinijev izrek

[ int_0 ^ 1 int _ {- 1} ^ 3 (3 - x + 4y) , dy , dx. neštevilno ]

b. Pokažite, da ( displaystyle 0 leq iint_R sin pi x , cos pi y , dA leq frac {1} {32} ) kjer je (R = levo (0, frac {1} {4} desno) levo ( frac {1} {4}, frac {1} {2} desno) ).

Namig

Uporabite lastnosti i. in ii. in ovrednotite ponovljeni integral, nato pa uporabite lastnost v.

Odgovorite

a. (26 )

b. Odgovori se lahko razlikujejo.

Kot smo že omenili, lahko pri uporabi pravokotnih koordinat dvojni integral nad območjem (R ), označenim z ( iint_R f (x, y) , dA ), zapišemo kot ( iint_R , f (x, y) , dx , dy ) ali ( iint_R , f (x, y) , dy , dx. ) Naslednji primer kaže, da so rezultati enaki, ne glede na vrstni red integracije, ki jo izberemo.

Primer ( PageIndex {6} ): ovrednotenje ponovljenega integrala na dva načina

Vrnimo se na funkcijo (f (x, y) = 3x ^ 2 - y ) iz primera 1, tokrat čez pravokotno območje (R = [0,2] krat [0,3] ). Uporabite Fubinijev izrek za ovrednotenje ( iint_R f (x, y) , dA ) na dva različna načina:

  1. Najprej se integrira glede na (y ) in nato še na (x );
  2. Najprej se integrira glede na (x ) in nato še na (y ).

Rešitev

Slika ( PageIndex {6} ) prikazuje delovanje izračuna na dva različna načina.

  1. Najprej se integrira glede na (y ) in nato integrira glede na (x ):

[ začeti {poravnati *} iint_R f (x, y) , dA & = int_ {x = 0} ^ {x = 2} int_ {y = 0} ^ {y = 3} (3x ^ 2 - y) , dy , dx [4pt]
& = int_ {x = 0} ^ {x = 2} levo ( int_ {y = 0} ^ {y = 3} (3x ^ 2 - y) , dy desno) , dx = int_ {x = 0} ^ {x = 2} levo [3x ^ 2y - frac {y ^ 2} {2} bigg | _ {y = 0} ^ {y = 3} desno] , dx [4pt]
& = int_ {x = 0} ^ {x = 2} levo (9x ^ 2 - frac {9} {2} desno) , dx = 3x ^ 3 - frac {9} {2} x bigg | _ {x = 0} ^ {x = 2} = 15. konec {poravnaj *} ]

  1. Najprej se integrira glede na (x ) in nato integrira glede na (y ):
    [ začeti {poravnati *} iint_R f (x, y) , dA & = int_ {y = 0} ^ {y = 3} int_ {x = 0} ^ {x = 2} (3x ^ 2 - y) , dx , dy [4pt]
    & = int_ {y = 0} ^ {y = 3} levo ( int_ {x = 0} ^ {x = 2} (3x ^ 2 - y) , dx desno) , dy [ 4pt]
    & = int_ {y = 0} ^ {y = 3} levo [x ^ 3 - xy bigg | _ {x = 0} ^ {x = 2} desno] dy [4pt]
    & = int_ {y = 0} ^ {y = 3} (8 - 2y) , dy = 8y - y ^ 2 bigg | _ {y = 0} ^ {y = 3} = 15. end { poravnaj *} ]

Analiza

Pri obeh zaporedjih integracije dvojni integral dobi odgovor (15 ). Ta odgovor bi si morda želeli razlagati kot prostornino v kubičnih enotah trdne snovi (S ) pod funkcijo (f (x, y) = 3x ^ 2 - y ) nad območjem (R = [0, 2] krat [0,3] ). Vendar ne pozabite, da interpretacija dvojnega integrala kot (nepodpisanega) zvezka deluje le, kadar je integrand (f ) nenegativna funkcija nad osnovnim območjem (R ).

Vadba ( PageIndex {3} )

Ocenite

[ int_ {y = -3} ^ {y = 2} int_ {x = 3} ^ {x = 5} (2 - 3x ^ 2 + y ^ 2) , dx , dy. neštevilno ]

Namig

Uporabite Fubinijev izrek.

Odgovorite

(- frac {1340} {3} )

V naslednjem primeru vidimo, da je dejansko koristno spremeniti vrstni red integracije za lažje računanje. V tem poglavju se bomo večkrat vrnili k tej ideji.

Primer ( PageIndex {7} ): Preklop vrstnega reda integracije

Razmislite o dvojnem integralu ( displaystyle iint_R x , sin (xy) , dA ) nad regijo (R = velika {(x, y) , | , 0 leq x leq pi, , 1 leq y leq 2 big } ) (slika ( PageIndex {7} )).

  1. Dvojni integral izrazite na dva različna načina.
  2. Analizirajte, ali je ovrednotenje dvojnega integrala na en način lažje kot na drug način in zakaj.
  3. Ocenite integral.
  1. ( Iint_R x , sin (xy) , dA ) lahko izrazimo na dva načina: najprej z integracijo glede na (y ) in nato z (x ); drugič z integracijo glede na (x ) in nato glede na (y ).
    [ iint_R x , sin (xy) , dA = int_ {x = 0} ^ {x = pi} int_ {y = 1} ^ {y = 2} x , sin (xy ) , dy , dx neštevilno ]
    Najprej integriraj glede na (y ).
    [= int_ {y = 1} ^ {y = 2} int_ {x = 0} ^ {x = pi} x , sin (xy) , dx , dy neštevilno ]
    Najprej integriraj glede na (x ).
  2. Če se želimo integrirati glede na y najprej in nato integriramo glede na (x ), vidimo, da lahko uporabimo substitucijo (u = xy ), ki daje (du = x , dy ). Torej je notranji integral preprosto ( int sin u , du ) in lahko omejitve spremenimo tako, da so funkcije (x ),

[ iint_R x , sin (xy) , dA = int_ {x = 0} ^ {x = pi} int_ {y = 1} ^ {y = 2} x , sin (xy ) , dy , dx = int_ {x = 0} ^ {x = pi} levo [ int_ {u = x} ^ {u = 2x} sin (u) , du desno] , dx. neštevilno ]

Vendar pa je za integracijo najprej glede na (x ) in nato za integracijo glede na (y ) potrebna integracija po delih za notranji integral, z (u = x ) in (dv = sin (xy) dx )

Potem (du = dx ) in (v = - frac { cos (xy)} {y} ), torej

[ iint_R x sin (xy) , dA = int_ {y = 1} ^ {y = 2} int_ {x = 0} ^ {x = pi} x sin (xy) , dx , dy = int_ {y = 1} ^ {y = 2} levo [- frac {x , cos (xy)} {y} bigg | _ {x = 0} ^ {x = pi} + frac {1} {y} int_ {x = 0} ^ {x = pi} cos (xy) , dx desno] , dy. neštevilno ]

Ker se vrednotenje zapleta, bomo izračunali le enostavnejše izračune, kar je očitno prva metoda.

  1. Na lažji način ocenite dvojni integral.

[ začeti {poravnati *} iint_R x , sin (xy) , dA & = int_ {x = 0} ^ {x = pi} int_ {y = 1} ^ {y = 2} x , sin (xy) , dy , dx [4pt]
& = int_ {x = 0} ^ {x = pi} levo [ int_ {u = x} ^ {u = 2x} sin (u) , du desno] , dx = int_ { x = 0} ^ {x = pi} levo [- cos u bigg | _ {u = x} ^ {u = 2x} desno] , dx [4pt]
& = int_ {x = 0} ^ {x = pi} (- cos 2x + cos x) , dx [4pt]
& = levo (- frac {1} {2} sin 2x + sin x desno) bigg | _ {x = 0} ^ {x = pi} = 0. konec {poravnaj *} ]

Vadba ( PageIndex {4} )

Izračunajte integral ( displaystyle iint_R xe ^ {xy} , dA ), kjer (R = [0,1] krat [0, ln 5] ).

Namig

Najprej integrirajte glede na (y ).

Odgovorite

( frac {4 - ln 5} { ln 5} )

Uporabe dvojnih integralov

Dvojni integrali so zelo koristni za iskanje območja območja, ki ga omejujejo krivulje funkcij. To situacijo podrobneje opisujemo v naslednjem poglavju. Če pa je območje pravokotne oblike, lahko njegovo območje najdemo z integracijo konstantne funkcije (f (x, y) = 1 ) nad regijo (R ).

Opredelitev: območje regije

Območje regije (R ) je podano z [A (R) = iint_R 1 , dA. ]

Ta opredelitev je smiselna, ker je z uporabo (f (x, y) = 1 ) in ocenjevanjem integrala rezultat dolžine in širine. Preverimo to formulo s primerom in poglejmo, kako to deluje.

Primer ( PageIndex {8} ): Iskanje območja z dvojnim integralom

Poiščite območje regije (R = big {, (x, y) , | , 0 leq x leq 3, , 0 leq y leq 2 big } ) po z uporabo dvojnega integrala, to je z integracijo (1 ) preko regije (R ).

Rešitev

Območje je pravokotno z dolžino (3 ) in širino (2 ), zato vemo, da je območje (6 ). Isti odgovor dobimo, če uporabimo dvojni integral:

[A (R) = int_0 ^ 2 int_0 ^ 3 1 , dx , dy = int_0 ^ 2 levo [x veliko | _0 ^ 3 desno] , dy = int_0 ^ 2 3 dy = 3 int_0 ^ 2 dy = 3y bigg | _0 ^ 2 = 3 (2) = 6 , besedilo {enote} ^ 2. Nešteto ]

Že smo videli, kako lahko z dvojnimi integrali poiščemo prostornino trdne snovi, omejene zgoraj s funkcijo (f (x, y) geq 0 ) nad območjem (R ), ki je predvideno (f (x, y) geq 0 ) za vse ((x, y) ) v (R ). Tu je še en primer za ponazoritev tega koncepta.

Primer ( PageIndex {9} ): Prostornina eliptičnega paraboloida

Poiščite prostornino (V ) trdne snovi (S ), ki je omejena z eliptičnim paraboloidom (2x ^ 2 + y ^ 2 + z = 27 ), ravninama (x = 3 ) in (y = 3 ) in tri koordinatne ravnine.

Rešitev

Najprej opazite graf površine (z = 27 - 2x ^ 2 - y ^ 2 ) na sliki ( PageIndex {8} ) (a) in nad kvadratnim območjem (R_1 = [-3,3 ] krat [-3,3] ). Vendar potrebujemo prostornino trdne snovi, ki jo omejujejo eliptični paraboloid (2x ^ 2 + y ^ 2 + z = 27 ), ravnini (x = 3 ) in (y = 3 ) ter tri koordinatne ravnine.

Zdaj pa si oglejmo graf površine na sliki ( PageIndex {8} ) (b). Prostornino (V ) določimo z vrednotenjem dvojnega integrala nad (R_2 ):

[ začeti {poravnati *} V & = iint_R z , dA = iint_R (27 - 2x ^ 2 - y ^ 2) , dA [4pt]
& = int_ {y = 0} ^ {y = 3} int_ {x = 0} ^ {x = 3} (27 - 2x ^ 2 - y ^ 2) , dx , dy & & text { Pretvori v dobesedni integral.} [4pt]
& = int_ {y = 0} ^ {y = 3} [27x - frac {2} {3} x ^ 3 - y ^ 2x] bigg | _ {x = 0} ^ {x = 3} , dy & & text {Integriraj glede na $ x $.} [4pt]
& = int_ {y = 0} ^ {y = 3} (63 - 3y ^ 2) dy = 63 y - y ^ 3 bigg | _ {y = 0} ^ {y = 3} = 162. konec {poravnaj *} ]

Vadba ( PageIndex {5} )

Poiščite prostornino trdne snovi, omejene zgoraj z grafom (f (x, y) = xy sin (x ^ 2y) ) in spodaj z ravnino (xy ) na pravokotnem območju (R = [0,1] krat [0, pi] ).

Namig

Grafizirajte funkcijo, nastavite integral in uporabite ponovljeni integral.

Odgovorite

( frac { pi} {2} )

Spomnimo se, da smo povprečno vrednost funkcije ene spremenljivke na intervalu ([a, b] ) opredelili kot

[f_ {ave} = frac {1} {b - a} int_a ^ b f (x) , dx. ]

Podobno lahko določimo povprečno vrednost funkcije dveh spremenljivk v regiji (R ). Glavna razlika je v tem, da delimo s površino namesto s širino intervala.

Opredelitev: POVPREČNA VREDNOST FUNKCIJE

The povprečna vrednost funkcije dveh spremenljivk v regiji (R ) je

[F_ {ave} = frac {1} { text {Območje} , R} iint_R f (x, y) , dx , dy. ]

V naslednjem primeru najdemo povprečno vrednost funkcije nad pravokotnim območjem. To je dober primer pridobivanja koristnih informacij za integracijo z izvedbo posameznih meritev prek mreže, namesto da bi poskušali najti algebrski izraz za funkcijo.

Primer ( PageIndex {10} ): Izračun povprečne nevihtne padavine

Vremenski zemljevid na sliki ( PageIndex {9} ) prikazuje nenavadno vlažen nevihtni sistem, povezan z ostanki orkana Karl, ki je septembra na nekaterih delih Srednjega zahoda septembra odvrgel 100–200 mm dežja. 22. – 23. 2010. Območje padavin je merilo 300 kilometrov vzhodno proti zahodu in 250 kilometrov severno proti jugu. Ocenite povprečno količino padavin na celotnem območju v teh dveh dneh.

Rešitev

Izvor postavite na jugozahodni kot zemljevida, tako da lahko vse vrednosti veljajo za prvi kvadrant in so zato vse pozitivne. Zdaj razdelite celoten zemljevid na šest pravokotnikov ((m = 2 ) in (n = 3) ), kot je prikazano na sliki ( PageIndex {9} ). Predpostavimo, da (f (x, y) ) označuje nevihtne padavine v palcih na točki približno (x ) milj vzhodno od izhodišča in (y ) milj severno od izhodišča. Naj (R ) predstavlja celotno površino (250 krat 300 = 75000 ) kvadratnih milj. Potem je površina vsakega podkotnika

[ Delta A = frac {1} {6} (75000) = 12500. nešteto ]

Predpostavimo, da so ((x_ {ij} *, y_ {ij} *) ) približno srednje točke vsakega podkotnika (R_ {ij} ). Na vsaki od teh točk si oglejte barvno označeno območje in ocenite količino padavin. Padavine na vsaki od teh točk lahko ocenimo kot:

  • Ob ( (x_ {11}, y_ {11} )) je količina padavin 0,08.
  • Ob ( (x_ {12}, y_ {12} )) je količina padavin 0,08.
  • Ob ( (x_ {13}, y_ {13} )) je količina padavin 0,01.
  • Ob ( (x_ {21}, y_ {21} )) je količina padavin 1,70.
  • Ob ( (x_ {22}, y_ {22} )) je količina padavin 1,74.
  • Ob ( (x_ {23}, y_ {23} )) je količina padavin 3,00.

Po naši definiciji je bila povprečna nevihtna količina padavin na celotnem območju v teh dveh dneh

[ začeti {poravnati *} f_ {ave} = frac {1} {Območje , R} iint_R & = f (x, y) , dx , dy = frac {1} {75000} iint_R f (x, y) , dx , dy [4pt]
& približno frac {1} {75000} sum_ {i = 1} ^ 3 sum_ {j = 1} ^ 2 f (x_ {ij} ^ *, y_ {ij} ^ *) Delta A [4pt]
& = frac {1} {75000} [f (x_ {11} ^ *, y_ {11} ^ *) Delta A + f (x_ {12} ^ *, y_ {12} ^ *) Delta A + f (x_ {13} ^ *, y_ {13} ^ *) Delta A + f (x_ {21} ^ *, y_ {21} ^ *) Delta A + f (x_ {22} ^ *, y_ {22} ^ *) Delta A + f (x_ {23} ^ *, y_ {23} ^ *) Delta A] [4pt]
& približno frac {1} {75000} [0,08 + 0,08 + 0,01 + 1,70 + 1,74 + 3,00] Delta A [4pt]
& = frac {1} {75000} [0,08 + 0,08 + 0,01 + 1,70 + 1,74 + 3,00] 12500 [4pt]
& = frac {1} {6} [0,08 + 0,08 + 0,01 + 1,70 + 1,74 + 3,00] [4pt] & približno 1,10 ; besedilo {v}. end {poravnaj *} ]

V obdobju med 22. in 23. septembrom 2010 je bilo na tem območju povprečno nevihta približno 1,10 palca.

Vadba ( PageIndex {6} )

Prikazan je konturni zemljevid za funkcijo (f (x, y) ) na pravokotniku (R = [-3,6] krat [-1, 4] ).

a. Uporabite pravilo srednje točke z (m = 3 ) in (n = 2 ), da ocenite vrednost ( displaystyle iint_R f (x, y) , dA. )

b. Ocenite povprečno vrednost funkcije (f (x, y) ).

Namig

Razdelite območje na šest pravokotnikov in s konturnimi črtami ocenite vrednosti za (f (x, y) ).

Odgovorite

Odgovori na oba dela a. in b. se lahko razlikujejo.

Ključni pojmi

  • Z dvojno Riemannovo vsoto lahko približamo prostornino trdne snovi, omejene zgoraj s funkcijo dveh spremenljivk v pravokotnem območju. Z določitvijo meje ta postane dvojni integral, ki predstavlja prostornino trdne snovi.
  • Lastnosti dvojnega integrala so uporabne za poenostavitev izračuna in iskanje meja njihovih vrednosti.
  • Fubinijev izrek lahko uporabimo za pisanje in vrednotenje dvojnega integrala kot ponovljenega integrala.
  • Dvojni integrali se uporabljajo za izračun površine regije, prostornine pod površino in povprečne vrednosti funkcije dveh spremenljivk v pravokotnem območju.

Ključne enačbe

  • [ iint_R f (x, y) , dA = lim_ {m, n rightarrow infty} sum_ {i = 1} ^ m sum_ {j = 1} ^ nf (x_ij *, y_ij *) , ΔA nešteto ]
  • [ int_a ^ b int_c ^ d f (x, y) , dx , dy = int_a ^ b levo [ int_c ^ d f (x, y) , dy desno] dx neštevilno ] ali

    [ int_c ^ d int_a ^ b f (x, y) , dx , dy = int_c ^ d levo [ int_a ^ b f (x, y) , dx desno] dy neštevilno ]

  • [f_ {ave} = frac {1} { text {Območje} , R} iint_R f (x, y) , dx , dy neštevilno ]

Slovarček

dvojni integral
funkcije (f (x, y) ) nad območjem (R ) v ravnini (xy ) - je opredeljena kot meja dvojne Riemannove vsote,
[ iint_R f (x, y) , dA = lim_ {m, n rightarrow infty} sum_ {i = 1} ^ m sum_ {j = 1} ^ nf (x_ {ij} ^ * , y_ {ij} ^ *) , Delta A. neštevilno ]
dvojna Riemannova vsota
funkcije (f (x, y) ) nad pravokotnim območjem (R ) je
[ sum_ {i = 1} ^ m sum_ {j = 1} ^ n f (x_ {ij} ^ *, y_ {ij} ^ *) , Delta A, nonumber ]
kjer je (R ) razdeljen na manjše podpravokotnike (R_ {ij} ) in ((x_ {ij} ^ *, y_ {ij} ^ *) ) je poljubna točka v (R_ {ij} )
Fubinijev izrek
če je (f (x, y) ) funkcija dveh spremenljivk, ki je neprekinjena v pravokotni regiji (R = velika {(x, y) v mathbb {R} ^ 2 , | ,a leq x leq b, , c leq y leq dig}), then the double integral of (f) over the region equals an iterated integral,
[displaystyleiint_R f(x,y) , dA = int_a^b int_c^d f(x,y) ,dx , dy = int_c^d int_a^b f(x,y) ,dx , dy onumber]
iterated integral
for a function (f(x,y)) over the region (R) is

a. (displaystyle int_a^b int_c^d f(x,y) ,dx , dy = int_a^b left[int_c^d f(x,y) , dy ight] , dx,)

b. (displaystyle int_c^d int_a^b f(x,y) , dx , dy = int_c^d left[int_a^b f(x,y) , dx ight] , dy,)

where (a,b,c), and (d) are any real numbers and (R = [a,b] imes [c,d])

Contributors and Attributions

  • Gilbert Strang (MIT) in Edwin "Jed" Herman (Harvey Mudd) s številnimi avtorji. Ta vsebina podjetja OpenStax je licencirana z licenco CC-BY-SA-NC 4.0. Prenesite ga brezplačno na http://cnx.org.


15.4 Pendulums

Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.

The Simple Pendulum

A simple pendulum is defined to have a point mass, also known as the pendulum bob , which is suspended from a string of length L with negligible mass (Figure 15.20). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if θ θ is less than about 15 ° . 15 ° . Even simple pendulum clocks can be finely adjusted and remain accurate.

Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

Example 15.3

Measuring Acceleration due to Gravity by the Period of a Pendulum

Strategy

Rešitev

Significance

An engineer builds two simple pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced by 12 ° 12 ° .

Physical Pendulum

Any object can oscillate like a pendulum. Consider a coffee mug hanging on a hook in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum until the oscillations die out. We have described a simple pendulum as a point mass and a string. A physical pendulum is any object whose oscillations are similar to those of the simple pendulum, but cannot be modeled as a point mass on a string, and the mass distribution must be included into the equation of motion.

As for the simple pendulum, the restoring force of the physical pendulum is the force of gravity. With the simple pendulum, the force of gravity acts on the center of the pendulum bob. In the case of the physical pendulum, the force of gravity acts on the center of mass (CM) of an object. The object oscillates about a point O. Consider an object of a generic shape as shown in Figure 15.21.

Using the small angle approximation and rearranging:

Once again, the equation says that the second time derivative of the position (in this case, the angle) equals minus a constant ( − m g L I ) ( − m g L I ) times the position. The solution is


Uvod

Functions with values in a (nonlinear) subset of a vector space appear in several applications of imaging and in inverse problems, e.g.,

Interferometric Synthetic Aperture Radar (InSAR) is a technique used in remote sensing and geodesy to generate, for example, digital elevation maps of the earth’s surface. InSAR images represent phase differences of waves between two or more SAR images, cf. [44, 53]. Therefore, InSAR data are functions (f:Omega ightarrow >^1subseteq >^2) . The pointwise function values are on the (>^1) , which is considered embedded into (>^2) .

A color image can be represented as a function in HSV space (hue, saturation, value) (see, e.g., [48]). Color images are then described as functions (f:Omega ightarrow K subseteq >^3) . Here (Omega ) is a plane in (>^2) , the image domain, and K (representing the HSV space) is a cone in three-dimensional space (>^3) .

Estimation of the foliage angle distribution has been considered, for instance, in [39, 51]. Therefore, the imaging function is from (Omega subset >^2) , a part of the Earth’s surface, into (mathbb ^2 subseteq >^3) , representing foliage angle orientation.

Estimation of functions with values in (SO(3) subseteq >^<3 imes 3>) . Such problems appear in Cryo-Electron Microscopy (see, for instance, [38, 58, 61]).

We emphasize that we are analyzing vector-, matrix-, tensor- valued functions, where pointwise function evaluations belong to some given (sub)set, but are always elements of the underlying vector space. This should not be confused with set-valued functions, where every function evaluation can be a set.

Inverse problems and imaging tasks, such as the ones mentioned above, might be unstable, or even worse, the solution could be ambiguous. Therefore, numerical algorithms for imaging need to be regularizing to obtain approximations of the desired solution in a stable manner. Consider the operator equation

where we assume that only (noisy) measurement data (v^delta ) of (v^0) become available. In this paper the method of choice is variational regularization which consists in calculating a minimizer of the variational regularization functional

w is an element of the set of admissible functions.

is an operator modeling the image formation process (except the noise).

(mathcal ) is called the podatkov ali fidelity term, which is used to compare a pair of data in the image domain, that is to quantify the difference of the two data sets.

(mathcal ) is called regularization functional, which is used to impose certain properties onto a minimizer of the regularization functional (mathcal ) .

(alpha > 0) is called regularization parameter and provides a trade off between stability and approximation properties of the minimizer of the regularization functional (mathcal ) .

(v^delta ) denotes measurement data, which we consider noisy.

(v^0) denotes the exact data, which we assume to be not necessarily available.

The main objective of this paper is to introduce a general class of regularization functionals for functions with values in a set of vectors. In order to motivate our proposed class of regularization functionals, we review a class of regularization functionals appropriate for analyzing intensity data.

Variational Regularization for Reconstruction of Intensity Data

Opposite to what we consider in the present paper, most commonly, imaging data v and admissible functions w, respectively, are considered to be representable as intensity functions. That is, they are functions from some subset (Omega ) of an Euclidean space with real values.

In such a situation, the most widely used regularization functionals use regularization terms consisting of powers of Sobolev (see [12, 15, 16]) or total variation semi-norms [54]. It is common to speak about Tikhonov regularization (see, for instance, [59]) when the data term and the regularization functional are squared Hilbert space norms, respectively. For the Rudin, Osher, Fatemi (ROF) regularization [54], also known as total variation regularization, the data term is the squared (L^2) -norm and (mathcal (w) = |w|_) is the total variation semi-norm. Nonlocal regularization operators based on the generalized nonlocal gradient are used in [35].

Other widely used regularization functionals are sparsity promoting [22, 41], Besov space norms [42, 46] and anisotropic regularization norms [47, 56]. Aside from various regularization terms, there also have been proposed different fidelity terms other than quadratic norm fidelities, like the p-th powers of (ell ^p) and (L^p) -norms of the differences of F(w) in v , [55, 57], maximum entropy [26, 28] and Kullback–Leibler divergence [52] (see [50] for some reference work).

Our work utilizes results from the seminal paper of Bourgain, Brézis and Mironescu [14], which provides an equivalent derivative-free characterization of Sobolev spaces and the space , the space of functions of bounded total variation, which consequently, in this context, was analyzed in Dávila and Ponce [23, 49], respectively. It is shown in [14, Theorems 2 and 3’] and [23, Theorem 1] that when (( ho _varepsilon )_) is a suitable sequence of nonnegative, radially symmetric, radially decreasing mollifiers, then

Hence, ( ilde<>>_varepsilon ) approximates powers of Sobolev semi-norms and the total variation semi-norm, respectively. Variational imaging, consisting in minimization of (mathcal ) from Eq. 1.2 with (>) replaced by ( ilde<>>_varepsilon ) , has been considered in [3, 11].

Regularization of Functions with Values in a Set of Vectors

In this paper we generalize the derivative-free characterization of Sobolev spaces and functions of bounded variation to functions (u:Omega ightarrow K) , where K is some set of vectors, and use these functionals for variational regularization. The applications we have in mind contain that K is a closed subset of (>^M) (for instance, HSV data) with nonzero measure, or that K is a submanifold (for instance, InSAR data).

The reconstruction of manifold-valued data with variational regularization methods has already been subject to intensive research (see, for instance, [4, 17,18,19, 40, 62]). The variational approaches mentioned above use regularization and fidelity functionals based on Sobolev and TV semi-norms: a total variation regularizer for cyclic data on (>^1) was introduced in [18, 19], see also [7, 9, 10]. In [4, 6] combined first- and second-order differences and derivatives were used for regularization to restore manifold-valued data. The later mentioned papers, however, are formulated in a finite-dimensional setting, opposed to ours, which is considered in an infinite-dimensional setting. Algorithms for total variation minimization problems, including half-quadratic minimization and nonlocal patch-based methods, are given, for example, in [4, 5, 8] as well as in [37, 43]. On the theoretical side the total variation of functions with values in a manifold was investigated by Giaquinta and Mucci using the theory of Cartesian currents in [33, 34], and earlier [32] if the manifold is (>^1) .

Content and Particular Achievements of the Paper

The contribution of this paper is to introduce and analytically analyze double integral regularization functionals for reconstructing functions with values in a set of vectors, generalizing functionals of the form Eq. 1.3. Moreover, we develop and analyze fidelity terms for comparing manifold-valued data. Summing these two terms provides a new class of regularization functionals of the form Eq. 1.2 for reconstructing manifold-valued data.

When analyzing our functionals, we encounter several differences to existing regularization theory (compare Sect. 2):

The admissible functions, where we minimize the regularization functional on, do form only a set but ne a linear space. As a consequence, well-posedness of the variational method (that is, existence of a minimizer of the energy functional) cannot directly be proven by applying standard direct methods in the Calculus of Variations [20, 21].

The regularization functionals are defined via metrics and not norms, see Sect. 3.

In general, the fidelity terms are non-convex. Stability and convergence results are proven in Sect. 4.

The model is validated in Sect. 6 where we present numerical results for denoising and inpainting of data of InSAR type.


Double Head Double Acting Series Hydraulic Pumps

A Double Airhead Double Acting reciprocating, positive displacement pump. Compact in size, minimal weight and ability to operate in many environments.

On This Page

Hydratron Double Airhead Double Acting Series, Air Driven Liquid Pump,

is a reciprocating, positive displacement pump with a directly coupled, integral linear reciprocating air motor designed to generate High-Pressure Liquid flow.

The pressure increase is made possible by means of the area ratio between the two larger diameter air drive pistons directly connected to two smaller diameter liquid plungers. Using our various material & seal options they are suitable for pumping a wide range of liquids including water, oil, water/glycol and many other chemical fluids.

All Models can easily be incorporated into easy to operate Standard System hydraulic power units.


Display Driver Uninstaller Download version 18.0.4.1

Download Display Driver Uninstaller DDU - Display Driver Uninstaller is a driver removal utility that can help you completely uninstall AMD/NVIDIA graphics card drivers and packages from your system, without leaving leftovers behind (including registry keys, folders and files, driver store).

The AMD/NVIDIA video drivers can normally be uninstalled from the Windows Control panel, this driver uninstaller program was designed to be used in cases where the standard driver uninstall fails, or anyway when you need to thoroughly delete NVIDIA and ATI video card drivers. The current effect after you use this driver removal tool will be similar as if its the first time you install a new driver just like a fresh, clean install of Windows. As with any tool of this kind, we recommend creating a new system restore point before using it, so that you can revert your system at any time if you run into problems.

If you have a problem installing an older driver or newer one, give it a try as there are some reports that it fix those problems. DDU is an application that is programmed by Ghislain Harvey aka Wagnard in our forums, Guru3D.com is the official download partner for this handy application.

Recommended usage

  • The tool can be used in Normal mode but for absolute stability when using DDU, Safemode is always the best.
  • Make a backup or a system restore (but it should normally be pretty safe).
  • It is best to exclude the DDU folder completely from any security software to avoid issues.

Keep note that NVIDIA/AMD did not have anything to do with this, I do not work at or for NVIDIA/AMD and they should not be held responsible for anything that may go wrong with this application.


As I always use Intel C/C++ and trying to check, I found a lot of tests with only (50..80) bits of precision, that was using /fp:fast2, and /fp:fast is almost the same.
Testing with /fp:strict the results are very similar to MS-C.
A test with /fp:precise returns more bits of accuracy that MS-C, with the only exception of fma() which had 86 versus 93 with MS-C.

Debugging the code line by line shows that both the 'unsigned long long' and the 'long long' constructors rely on undefined behaviour when converting extremely large values, namely conversion of a value outside the '(unsigned) long long' range to '(unsigned) long long'. In the case of 'long long', this conversion is benign. In the case of 'unsigned long long', the conversion gives an error.

Replacing the constructors for '(unsigned) long long' with the following code will work for all values, assuming that:

1. The processor uses twos-complement arithmetic
2. The '(unsigned) long long' type has no more than twice the number of bits in a 'double'

The code separates an '(unsigned) long long' type into two parts that may be represented exactly within a 'double'. It then adds the two 'double's to give a normalized result. It has been tested with values LLONG_MIN, LLONG_MAX, and ULLONG_MAX, and gives the correct results for all three operands.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

My fix (in C# where a ulong is an unsigned long long in C++) is as follows:

This may be a kludge as perhaps normalizing can be done to speed things up a bit but I'm out of my depth there. Same is true for signed long long.
Thinking about it I suspect that the original designers of QD (Bailey, et al) intentionally created the code this way - there is simply a recognised loss of precision. But my thoughts are that since double-doubles accommodate 31 digits precision then they should accommodate long longs accurately.

Apologies for the delay I was away from my computer.

I am testing solutions for all the problems that you raised, and will have a new version (1.1.3) of the package on the website later tonight Israel time (UTC+2).

The problem with unsigned long long is mine alone Bailey et al. did not have a constructor for (unsigned) long long types. Somehow, the subtraction operation got reversed:

For |h| < 2^53, x_[1] == 0, so the order of subtraction is unimportant. I did not test my code for very large numbers, and so missed the bug.

My thanks again for bringing these bugs to my attention.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

I'm sorry my test code had an error, making me think that there was a problem. I erased my message, but it had apparently already been sent to you.

Apologies for the false alarm.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

I executed the following code (results in comments):

I've fixed the code (in C# as I'm transcoding the C++) as follows but it's not completely tested (NB. the method is a member of struct DDReal):

I can confirm the error. It is due to the fact that 9.9. 92 is converted to <10.0, > while 9.9. 91 is converted to <9.9. >.

A workaround is to use the trunc() function and cast the result to int. I will see if I can create a better fix if not, the toInt() method will be updated to call the trunc() function.

Note that similar issues exist in the toLong() in toLongLong() methods.

My thanks for discovering this.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

No, this is not an error. The double-double type represents numbers as a sum of two doubles a+b, where b < 0.5×eps×a. There is no lower bound on b, so any number of the form 1+2^n can be represented. For example, 1+2^-300 is representative as a double-double (a=1, b=2^-300), but is not representable as a binary128.

The double-double type provides at least 106 bits of precision, so epsilon in numeric_limits is set to this value.

This "wobbling precision" is one of the differences between true ieee arithmetic and the double-double type.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

dd_real a = "123456789012345678901234567890"
a *= 1e-35
std::string s = a.to_string()
std::cout << s << " "

No, my mistake - there's no error. It seems that the double 1e-35 can,t be being represented exactly in binary (correct?) hence it's invisible trailing digits are propogated.

Doing:
dd_real a = "123456789012345678901234567890"
dd_real b = "1e-35"
a *= b
yields the correct answer.

You are correct in that 1e-35 cannot be represented exactly in binary. Your second solution (setting b = "1e-35") gave a more accurate representation - 106 bits, rather than 53 - but still did not give an infinitely accurate value.

It is this sort of problem that makes floating-point arithmetic such a joy and a challenge.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

  1. AFAIK, It's supported only on Intel processors
  2. Even on those processors, it's supported only by the old-style 80x87 instructions, not the SSE instructions

This uniformity does not exist for 'long double'. The Intel processors have limited support in hardware for an 80-bit extended type, PowerPC processors typically implement 'long double' in software as a double-double type, while some SPARC processors implement it in software as a 128-bit quadruple-precision type. These processors could use 'long double' as an alternative to 'double-double', but you then run into the same problems we had pre-IEEE 754: running the same program on different processors does not necessarily give the same result!

Until IEEE 754 binary128 is natively supported on all processors, I submit that with all its faults, 'double-double' is the best choice for portability and high accuracy.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

In your case, a == -1.32, b == 2.0, which leads to int_b == 2.0, frac_b == 0.0.

The problem is that if a < 0.0, log( a ) cannot be calculated. However, if frac_b == 0.0, this whole line should be skipped.

The code may be fixed by replacing the above with

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

Thanks for the quick reply.

Also I noticed that 0/0 gives inf. It should yield nan.

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

General News Suggestion Question Bug Answer Joke Praise Rant Admin

Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages.


Best paid email clients:

1. Microsoft Outlook

Microsoft&rsquos classic email client

Reasons to buy

Microsoft&rsquos Outlook is the de facto email client for most businesses and enterprises, and has been around for decades, with its origins dating back to MS-DOS. Obviously it has tight integration with other Microsoft services, and that takes email beyond the simple exchange of messages.

Outlook has the advantage of being fully integrated with the Outlook Calendar, making it a snap to share calendars to coordinate meetings. This integration also extends to Outlook Contacts. Outlook is supported for the Windows platform, but also across the mobile platforms of iOS and Android as well.

Microsoft Outlook is available as part of the Microsoft Office suite, which can be purchased as the standalone Office 2016, or the subscription-based Microsoft 365.

2. eM Client

A full-featured alternative email client

Reasons to buy

eM Client has been around for nearly 10 years now, and throughout that long development it's evolved into the best alternative email client for Windows.

It offers a wide array of features, including a calendar, contacts and chat. Support is provided for all the major email services including Gmail, Yahoo, iCloud and Outlook.com. The latest version also offers PGP encryption, live backup, basic image editing capabilities and auto-replies for Gmail.

There is a free tier, but you need the Pro version for commercial use, and that also gives you VIP support and unlimited accounts (the free product is limited to two email accounts). The Pro version has a one-time license fee.

eM Client makes it easy to migrate your messages from Gmail, Exchange, iCloud and Outlook.com &ndash just enter your email address and the client will adjust the appropriate settings for you. eM Client can also import your contacts and calendar, and it's easy to deselect these options if you'd prefer to manage them separately.

There's an integrated chat app too, with support for common platforms including Jabber and Google Chat, and the search function is far superior to those you'll find in webmail interfaces.

3. Mailbird

The email client that bristles with app integrations

Reasons to buy
Reasons to avoid

Mailbird is an email client that promises to &ldquosave time managing multiple accounts,&rdquo and to make your email &ldquoeasy and beautiful&rdquo. It comes in two main versions: Personal and Business.

While beauty may be in the eye of the beholder, as they say, it&rsquos undeniable that Mailbird Business offers many free themes to make email a more enjoyable and customizable experience.

Unlike some more Microsoft-centric email clients, Mailbird Business supports a diverse range of integrated apps, including WhatsApp, Google Docs, Google Calendar, Facebook, Twitter, Dropbox and Slack, all making for a better streamlined workflow. However, one downside to bear in mind here is that there&rsquos no support for filters or rules to organize your inbox.

Mailbird Personal is available for free, with Mailbird Business available as a subscription or a one-time lifetime license.

4. Inky

The anti-phishing email client

Reasons to buy
Reasons to avoid

Inky is an email client that focuses on security, using AI and machine learning algorithms to block all manner of phishing attacks which might otherwise get through.

This client uses an &lsquoInky Phish Fence&rsquo that scans both internal and external emails to flag phishing attempts. The proprietary machine learning technology can literally read an email to determine if it has phishing content, and then is able to quarantine the email, or deliver it with the malicious links disabled. It also takes things a step further and offers an analytics dashboard, which allows an administrator to see patterns of attacks based on dates, or targeted users.

The Inky email client does offer a free trial, but sadly, pricing details aren&rsquot made available on the Inky website. However, the site does note that pricing is per mailbox per month on a subscription, with volume discounts available.

5. Hiri

Packed with time-saving tools that'll improve email habits

Reasons to buy

Hiri is a paid-for premium email client that is designed primarily with business users in mind (it currently only supports Microsoft email services including Hotmail, Outlook and Exchange), but home users will also appreciate its productivity-boosting features.

If you find yourself spending too long managing, reading and replying to emails, Hiri is the email client for you. It includes a smart dashboard that lets you see how many unread messages you have at a glance and how long you should wait before checking them (after all, how many really need an instant reply?)

The Compose window is designed to save you time too, offering only the essential options (no fancy formatting) and including the subject line at the bottom so you don't have to write it until you know how to summarize the message.

These little touches make Hiri a truly exceptional client. If Microsoft is your email provider of choice, it should be well up your list. Hiri is available to buy annually or via a lifetime license for one-time fee. Both options offer a 7-day free trial.


For the past 12 years, GR8Conf has provided a high-quality conference experience for the tight-knit Apache Groovy programming language community. This year, we are joining forces with JDK IO - an annual conference run by the Danish Java user group covering technologies relevant to the entire JVM. The combined conference will be known as GR8Conf & JDK IO and will focus on All Things Groovy and Java, with DevOps, Microservices and Frontend Technologies sprinkled in.

We feature the Hackergarten and have an awesome Meet & Greet with craft beers brewed by the crew!


Ready to ki$ pa$words g00dby3? Soon, you'll be able to do just that with Duo.

Passwordless authentication promises to provide a frictionless login experience, while reducing administrative burden and overall security risks for your organization. In essence, it's a simpler, more secure way to MFA.

Our passwordless authentication solution is flexible and easy to set up, and it's designed with the same best-in-class usability you'd expect from any Duo product. Sign up today to receive updates on our upcoming passwordless experience.


News and Events

Catch up with GTK development

Get in touch with GTK developers through IRC. Get daily updates about GTK and its community from GTK blog or through its Twitter account.

Meet the community

As regularly as possible, GTK team meetings take place at conferences and hackfests to discuss the future of GTK and define a roadmap.

Contribute to GTK

If you are a developer and want to contribute to GTK, you are more than welcome to do so.


Poglej si posnetek: MAT120171211 Aplikácie určitého integrálu I. (December 2021).