EngComp e Prod 2018 1 Calculo II MCA502 P2 GABARITO
3 pág.

EngComp e Prod 2018 1 Calculo II MCA502 P2 GABARITO


Disciplina<strong>cálculo</strong>36 materiais5 seguidores
Pré-visualização1 página
1 
 
AVALIAÇÃO PRESENCIAL 
CADERNO DE PERGUNTAS 
 
curso: 
Engenharia de Computação/ 
Engenharia de Produção 
bimestre: 4º bimestre ano: 2018 | 2sem 
CÓDIGO DA PROVA 
P2 
 
\uf0b7 Preencha atentamente o cabeçalho de TODAS AS FOLHAS DE RESPOSTA que você utilizar. 
\uf0b7 Ao término da prova, entregue apenas a folha de resposta ao aplicador. Leve este caderno de 
perguntas consigo. 
Boa prova! 
 
disciplina: MCA502 \u2013 Cálculo II 
 
\uf0b7 É permitido o uso de formulário impresso. Não é permitido o uso de calculadora. 
 
Questão 1 (2 pontos) 
Sobre o campo vetorial \ufffd\u20d7\ufffd(\ud835\udc65,\ud835\udc66) = (
1
\ud835\udc66
+ \ud835\udc65\ud835\udc66 \u2212 2\ud835\udc65\ud835\udc66sen\u2061(\ud835\udc652\ud835\udc66))\ud835\udc56 + (
\u2212\ud835\udc65
\ud835\udc662
+ \ud835\udc652 \u2212 \ud835\udc652\ud835\udc60\ud835\udc52\ud835\udc5b(\ud835\udc652\ud835\udc66))\ud835\udc57 é correto afirmar que: 
 
a) o campo não é conservativo, pois \ud835\udc5f\ud835\udc5c\ud835\udc61\ufffd\u20d7\ufffd \u2260 0\u20d7\u20d7. 
b) \ud835\udc5f\ud835\udc5c\ud835\udc61\ufffd\u20d7\ufffd = 0\u20d7\u20d7\u2061e o campo não é conservativo. 
c) o campo é conservativo e \ud835\udf11(\ud835\udc65, \ud835\udc66) =
\ud835\udc65
\ud835\udc66
+ \ud835\udc652y + cos(\ud835\udc652\ud835\udc66) é uma função potencial. 
d) o campo é conservativo e \ud835\udf11(\ud835\udc65, \ud835\udc66) =
\ud835\udc65
\ud835\udc66
+ \ud835\udc652y \u2212 cos(\ud835\udc652\ud835\udc66) é uma função potencial. 
e) nenhuma das outras alternativas. 
 
 
Questão 2 (2 pontos) 
Calcule a massa da placa plana descrita por \ud835\udc652 +\ud835\udc662 \u2264 4, \ud835\udc66 \u2265 0, \ud835\udc65 \u2265 0, com densidade \ud835\udeff = \ud835\udeff(\ud835\udc65, \ud835\udc66) =
\u221a\ud835\udc652 + \ud835\udc662. 
 
a) \ud835\udf0b 
b) 
2\ud835\udf0b
3
 
c) 
8\ud835\udf0b
3
 
d) 
3\ud835\udf0b
2
 
e) Nenhuma das outras alternativas. 
 
 
Questão 3 (3 pontos) 
Calcule \u222c \ud835\udc52\ud835\udc66
3
\u221a2\ud835\udc67 +13
3
\ud835\udc51\ud835\udc66 \u2227 \ud835\udc51\ud835\udc67 + \ud835\udc66\ud835\udc672\ud835\udc51\ud835\udc67 \u2227 \ud835\udc51\ud835\udc65 +\u2061 ln(\ud835\udc652 + \ud835\udc664 + 1) \ud835\udc51\ud835\udc65 \u2227 \ud835\udc51\ud835\udc66\u2061\ud835\udc46 através da superfície S que é o 
bordo do elipsoide 
\ud835\udc652
9
+ \ud835\udc662 +
\ud835\udc672
4
\u2264 1,\u2061orientada com a normal que aponta para fora. 
 
Questão 4 (3 pontos) 
Calcule a massa do pedaço do cilindro \ud835\udc652+ \ud835\udc662 = 4 , \ud835\udc66 \u2265 0,\u2061acima do plano \ud835\udc67 = \u22122 e abaixo da superfície \ud835\udc67 =
4 \u2212 \ud835\udc66, com densidade \ud835\udeff = \ud835\udeff(\ud835\udc65, \ud835\udc66, \ud835\udc67) = 2\ud835\udc66 . 
 
2 
 
GABARITO 
curso: 
Engenharia de Computação/ 
Engenharia de Produção 
bimestre: 4º bimestre P2 
 
Questão 1 
Alternativa A. 
Como o campo é plano, segue que: 
 \ud835\udc5f\ud835\udc5c\ud835\udc61\ufffd\u20d7\ufffd = (
\ud835\udf15\ud835\udc44
\ud835\udf15\ud835\udc65
\u2212
\ud835\udf15\ud835\udc43
\ud835\udf15\ud835\udc66
) \ufffd\u20d7\u20d7\ufffd = (
\u22121
\ud835\udc662
+ 2\ud835\udc65 \u2212 2\ud835\udc65\ud835\udc60\ud835\udc52\ud835\udc5b(\ud835\udc652\ud835\udc66) \u2212 2\u2061\ud835\udc653\ud835\udc66\ud835\udc50\ud835\udc5c\ud835\udc60(\ud835\udc652\ud835\udc66) \u2212 (
\u22121
\ud835\udc662
+ \ud835\udc65 \u2212 2\ud835\udc65sen\u2061(\ud835\udc652\ud835\udc66) \u2212 2\u2061\ud835\udc653\ud835\udc66\ud835\udc50\ud835\udc5c\ud835\udc60(\ud835\udc652\ud835\udc66)))\ufffd\u20d7\u20d7\ufffd = 
\ud835\udc65\ufffd\u20d7\u20d7\ufffd \u2260 0\u20d7\u20d7 . Logo o campo não é conservativo. 
 
 
Questão 2 
Alternativa E. 
Massa = \u222c \ud835\udeff(\ud835\udc65, \ud835\udc66)\ud835\udc51\ud835\udc34\ud835\udc37 = \u222c \u221a\ud835\udc65
2 + \ud835\udc662\ud835\udc51\ud835\udc34\ud835\udc37 . 
Vamos usar coordenadas polares no cálculo da integral dupla. 
Parametrização: 
\ud835\udc65 = \ud835\udc5f\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 0 \u2264 \ud835\udf03 \u2264
\ud835\udf0b
2
 , (\ud835\udc66 \u2265 0, \ud835\udc65 \u2265 0) 
\ud835\udc66 = \ud835\udc5f\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 0 \u2264 \ud835\udc5f \u2264 2 
Jacobiano = \ud835\udc5f. 
\u222c \u221a\ud835\udc652+ \ud835\udc662\ud835\udc51\ud835\udc34\ud835\udc37 = \u222b \u222b \ud835\udc5f \u2219 \ud835\udc5f\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 = \u222b
1
3
\ud835\udc5f3|
2
0
\ud835\udc51\ud835\udf03 =
\ud835\udf0b
2
0
2
0
\ud835\udf0b
2
0 
= \u222b
8
3
\ud835\udf0b
2
0 \ud835\udc51\ud835\udf03 =
8
3
\u2219
\ud835\udf0b
2
=
4\ud835\udf0b
3
. 
 
 
Questão 3 
Pelo Teorema de Gauss vale: 
 \u222c \ud835\udc52\ud835\udc66
3
\u221a2\ud835\udc67 + 13
3
\ud835\udc51\ud835\udc66 \u2227 \ud835\udc51\ud835\udc67 + \ud835\udc66\ud835\udc672\ud835\udc51\ud835\udc67 \u2227 \ud835\udc51\ud835\udc65 + \u2061ln(\ud835\udc652+ \ud835\udc664 + 1) \ud835\udc51\ud835\udc65 \u2227 \ud835\udc51\ud835\udc66\u2061\ud835\udc46 = 
 =\u222d (
\ud835\udf15\ud835\udc43
\ud835\udf15\ud835\udc65
+
\ud835\udf15\ud835\udc44
\ud835\udf15\ud835\udc66
+
\ud835\udf15\ud835\udc45
\ud835\udf15\ud835\udc67\ud835\udc49
)\ud835\udc51\ud835\udc49 = \u222d \ud835\udc672\ud835\udc51\ud835\udc49\ud835\udc49 
Usamos coordenadas esféricas adaptadas ao elipsoide: 
\ud835\udc65 = 3 \ud835\udf0c\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 0 \u2264 \ud835\udf03 \u2264 2\ud835\udf0b 
\ud835\udc66 = \ud835\udf0c\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 0 \u2264 \ud835\udf0c \u2264 1 
\ud835\udc67 = 2\ud835\udf0c\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf11 0 \u2264 \ud835\udf11 \u2264 \ud835\udf0b 
Jacobiano = 3 \u2219 1 \u2219 2\ud835\udf0c2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11 = 6\ud835\udf0c2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11 . 
\u222d \ud835\udc672\ud835\udc51\ud835\udc49\ud835\udc49 = \u222b \u222b \u222b 4\ud835\udf0c
2\ud835\udc50\ud835\udc5c\ud835\udc602\ud835\udf11 \u2219 6\ud835\udf0c2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11\ud835\udc51\ud835\udf0c\ud835\udc51\ud835\udf11\ud835\udc51\ud835\udf03 =
1
0
\ud835\udf0b
0
2\ud835\udf0b
0 
= 24 \u2219 2\ud835\udf0b \u222b \ud835\udc50\ud835\udc5c\ud835\udc602\ud835\udf11\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf11
1
5
\ud835\udf0c5|
1
0
\ud835\udc51\ud835\udf11 =
\ud835\udf0b
0
48\ud835\udf0b
5
\u2219
1
3
(\u2212\ud835\udc50\ud835\udc5c\ud835\udc603\ud835\udf11)|
\ud835\udf0b
0
=
96\ud835\udf0b
15
=
32\ud835\udf0b
5
 
 
 
disciplina: MCA502 \u2013 Cálculo II 
3 
 
Questão 4 
Massa = \u222c \ud835\udeff(\ud835\udc65, \ud835\udc66, \ud835\udc67)\ud835\udc51\ud835\udc34\ud835\udc37 = \u222c 2\ud835\udc66\ud835\udc51\ud835\udc34\ud835\udc37 
Parametrização: 
\ud835\udc65 = 2\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 0 \u2264 \ud835\udf03 \u2264 \ud835\udf0b , pois \u2061\ud835\udc66 \u2265 0 
\ud835\udc66 = 2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 \u22122 \u2264 \ud835\udc67 \u2264 4 \u2212 2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 
\ud835\udc67 = \ud835\udc67 \ud835\udc51\ud835\udc34 = 2\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 
\u222c 2\ud835\udc66\ud835\udc51\ud835\udc34\ud835\udc37 = \u222b \u222b 4\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 \u2219 2\ud835\udc51\ud835\udc67\ud835\udc51\ud835\udf03 = \u222b 8\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 \u2219 \ud835\udc67|
4 \u2212 2\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03
\u22122
\ud835\udc51\ud835\udf03 =
\ud835\udf0b
0
4\u22122\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03
\u22122
\ud835\udf0b
0 
= \u222b (32\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 \u2212 16\ud835\udc60\ud835\udc52\ud835\udc5b2\ud835\udf03
\ud835\udf0b
0 +16\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03)\ud835\udc51\ud835\udf03 = \u222b (48\ud835\udc60\ud835\udc52\ud835\udc5b\ud835\udf03 \u2212 16(
1\u2212\ud835\udc50\ud835\udc5c\ud835\udc602\ud835\udf03
2
\ud835\udf0b
0 ))\ud835\udc51\ud835\udf03 
 
(\u221248\ud835\udc50\ud835\udc5c\ud835\udc60\ud835\udf03 \u2212 8\ud835\udf03 + 4\ud835\udc60\ud835\udc52\ud835\udc5b(2\ud835\udf03))|
\ud835\udf0b
0
= 48 \u2212 8\ud835\udf0b + 0 \u2212 (\u221248 \u2212 0 +0) = 96 \u2212 8\ud835\udf0b