Calcule a integral $\int_{C}\left(x^{2}+i y^{3}\right) d z$ onde $C$ é o segmento de reta de $z=1$ à $z=i$.
Passo 1
Beleza, vamos resolver essa integral
. Aqui estamos lidando com cálculo de integrais de linha no plano complexo. Para resolver isso, precisamos parametrizar o caminho
que vai de
a
, e então usar a parametrização para transformar a integral em uma integral no intervalo de 0 a 1.
Passo 2
Primeiro, vamos parametrizar o caminho
. Sabemos que
é uma reta que vai de
a
. Podemos escrever essa parametrização como
, onde
varia de 0 a 1. Aqui,
corresponde a
e
a
.
Passo 3
Agora, precisamos encontrar
, que é a derivada de
em relação a
. Derivando
, temos:
Passo 4
Também precisamos expressar
e
em termos de
. Sabendo que
, e usando a parametrização
, temos:
Passo 5
Substituímos
e
na função integranda
:
Passo 6
Agora, substituímos tudo na integral original:
Passo 7
Vamos expandir a expressão dentro da integral:
Distribuindo:
Simplificando:
Passo 8
Agora, integramos cada termo separadamente de
a
:
Calculando cada integral:
Passo 9
Agora, somamos todas as integrais:
Simplificando:
Para somar as frações, encontramos o denominador comum, que é 12:
Resposta
A integral
vale
.
Exercícios relacionados
Calcule a integral de linha, em que
é a curva espacial fornecida.
,
Calcule a integral de linha, em que
é a curva espacial fornecida.
,
C é o segmento de reta que vai de a
C é o segmento de reta que vai de
<html><head></head><body><div>Determine o módulo de: <span class="math-inline ">
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612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z”></path></g><g data-mml-node=“mo” transform=“translate(500,0)”><path data-c=“2212” d=“M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z”></path></g><g data-mml-node=“TeXAtom” data-mjx-texclass=“ORD” transform=“translate(1278,0)”><g data-mml-node=“mi”><path data-c=“1D422” d=“M72 610Q72 649 98 672T159 695Q193 693 217 670T241 610Q241 572 217 549T157 525Q120 525 96 548T72 610ZM46 442L136 446L226 450H232V62H294V0H286Q271 3 171 3Q67 3 49 0H40V62H109V209Q109 358 108 362Q103 380 55 380H43V442H46Z”></path></g></g></g><g data-mml-node=“mrow” transform=“translate(220,-376.9) scale(0.707)”><g data-mml-node=“mo”><path data-c=“28” d=“M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 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</body></html>
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612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z”></path></g><g data-mml-node=“mo” transform=“translate(500,0)”><path data-c=“2212” d=“M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z”></path></g><g data-mml-node=“TeXAtom” data-mjx-texclass=“ORD” transform=“translate(1278,0)”><g data-mml-node=“mi”><path data-c=“1D422” d=“M72 610Q72 649 98 672T159 695Q193 693 217 670T241 610Q241 572 217 549T157 525Q120 525 96 548T72 610ZM46 442L136 446L226 450H232V62H294V0H286Q271 3 171 3Q67 3 49 0H40V62H109V209Q109 358 108 362Q103 380 55 380H43V442H46Z”></path></g></g></g><g data-mml-node=“mrow” transform=“translate(220,-376.9) scale(0.707)”><g data-mml-node=“mo”><path data-c=“28” d=“M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 -226Q180 -141 137 -14T94 250Z”></path></g><g data-mml-node=“mn” transform=“translate(389,0)”><path data-c=“36” d=“M42 313Q42 476 123 571T303 666Q372 666 402 630T432 550Q432 525 418 510T379 495Q356 495 341 509T326 548Q326 592 373 601Q351 623 311 626Q240 626 194 566Q147 500 147 364L148 360Q153 366 156 373Q197 433 263 433H267Q313 433 348 414Q372 400 396 374T435 317Q456 268 456 210V192Q456 169 451 149Q440 90 387 34T253 -22Q225 -22 199 -14T143 16T92 75T56 172T42 313ZM257 397Q227 397 205 380T171 335T154 278T148 216Q148 133 160 97T198 39Q222 21 251 21Q302 21 329 59Q342 77 347 104T352 209Q352 289 347 316T329 361Q302 397 257 397Z”></path></g><g data-mml-node=“mo” transform=“translate(889,0)”><path data-c=“2B” d=“M56 237T56 250T70 270H369V420L370 570Q380 583 389 583Q402 583 409 568V270H707Q722 262 722 250T707 230H409V-68Q401 -82 391 -82H389H387Q375 -82 369 -68V230H70Q56 237 56 250Z”></path></g><g data-mml-node=“mn” transform=“translate(1667,0)”><path data-c=“32” d=“M109 429Q82 429 66 447T50 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</body></html>
Use o teorema de Cauchy-Goursat para mostrar que
, onde
é um contorno fechado envolvendo a origem uma vez no sentido anti-horário.