EngProdComp 2017 MCA502CalculoII P3 GABARITO
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EngProdComp 2017 MCA502CalculoII P3 GABARITO


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1 
 
AVALIAÇÃO PRESENCIAL 
CADERNO DE PERGUNTAS 
curso: Engenharia de Produção/ Engenharia de Computação bimestre: 4
o bimestre ano: 2018 | 1sem P3 
\u2022 Preencha atentamente o cabeçalho de TODAS AS FOLHAS DE RESPOSTA que você utilizar. 
\u2022 Ao término da prova, entregue apenas a folha de resposta ao aplicador. Leve este caderno de 
perguntas consigo. 
Boa prova! 
 
disciplina MCA502 - Cálculo II 
 
Questão 1 (1,0 ponto) 
A temperatura de uma placa plana é dada por \ud835\udc47\ud835\udc47 = \ud835\udc47\ud835\udc47(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) onde x e y são as coordenadas de um ponto 
genérico da placa. Um besouro anda sobre a placa seguindo uma trajetória \ud835\udc65\ud835\udc65 = \ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61) e \ud835\udc66\ud835\udc66 = \ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61), onde t é o 
tempo em segundos. Sabe-se que quando \ud835\udc61\ud835\udc61 = 5 o besouro está na posição (2,4) e que \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(5) = \u22121, \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(5) =2, \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(2,4) = 3 e \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(2,4) = 2. Podemos afirmar: 
a) No instante t = 5 a temperatura no besouro está aumentando 2 unidades por segundo. 
b) No instante t = 5 a temperatura no besouro está diminuindo 2 unidades por segundo. 
c) No instante t = 5 a temperatura no besouro está aumentando 1 unidade por segundo. 
d) No instante t = 5 a temperatura no besouro está diminuindo 1 unidades por segundo. 
e) Nenhuma das anteriores. 
 
Questão 2 (1,0 ponto) 
A massa do retângulo \ud835\udc45\ud835\udc45 = [2, 4]\ud835\udc4b\ud835\udc4b[0, 3] com densidade \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ud835\udc66\ud835\udc662 é: 
a) 18 
b) 54 
c) 9 
d) 6 
e) Nenhuma das anteriores. 
 
Questão 3 (1,0 ponto) 
Calcule a massa do arco de hélice \u210e(\ud835\udc61\ud835\udc61) = (3\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, 3\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, 4\ud835\udc61\ud835\udc61), 0 \u2264 \ud835\udc61\ud835\udc61 \u2264 \ud835\udf0b\ud835\udf0b
2
 com densidade \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67) = 4\ud835\udc67\ud835\udc67. 
a) 10\ud835\udf0b\ud835\udf0b2 
b) 10\ud835\udf0b\ud835\udf0b 
c) 20\ud835\udf0b\ud835\udf0b2 
d) 15\ud835\udf0b\ud835\udf0b2 
e) Nenhuma das anteriores. 
 
Questão 4 (1,0 ponto) 
A equação da reta tangente à curva \ud835\udefe\ud835\udefe(\ud835\udc61\ud835\udc61) = (\ud835\udc61\ud835\udc613, 5\ud835\udc61\ud835\udc612, \ud835\udc61\ud835\udc61) no ponto \ud835\udefe\ud835\udefe(2) é: 
a) \ud835\udc4b\ud835\udc4b(\ud835\udc50\ud835\udc50) = (8 \u2212 12\ud835\udc50\ud835\udc50, 20 \u2212 20\ud835\udc50\ud835\udc50, 2 + \ud835\udc50\ud835\udc50), \ud835\udc50\ud835\udc50 \u2208 \u211d 
b) \ud835\udc4b\ud835\udc4b(\ud835\udc50\ud835\udc50) = (8 + 12\ud835\udc50\ud835\udc50, 20 + 20\ud835\udc50\ud835\udc50, 2 + \ud835\udc50\ud835\udc50), \ud835\udc50\ud835\udc50 \u2208 \u211d 
c) \ud835\udc4b\ud835\udc4b(\ud835\udc50\ud835\udc50) = (8 \u2212 12\ud835\udc50\ud835\udc50, 20 + 20\ud835\udc50\ud835\udc50, 2 + \ud835\udc50\ud835\udc50), \ud835\udc50\ud835\udc50 \u2208 \u211d 
d) \ud835\udc4b\ud835\udc4b(\ud835\udc50\ud835\udc50) = (8 + 12\ud835\udc50\ud835\udc50, 20 \u2212 20\ud835\udc50\ud835\udc50, 2 + \ud835\udc50\ud835\udc50), \ud835\udc50\ud835\udc50 \u2208 \u211d 
e) Nenhuma das anteriores. 
 
 
CÓDIGO DA PROVA 
2 
 
Questão 5 (1,0 ponto) 
Calcule o trabalho realizado pelo campo vetorial \ufffd\u20d7\ufffd\ud835\udc39(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67) = 2\ud835\udc66\ud835\udc66\ud835\udea4\ud835\udea4 + \ud835\udc65\ud835\udc652\ud835\udea5\ud835\udea5 + \ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc67\ud835\udc58\ud835\udc58\ufffd\u20d7 
sobre a trajetória \ud835\udefe\ud835\udefe(\ud835\udc61\ud835\udc61) = (2\ud835\udc61\ud835\udc61, 3\ud835\udc61\ud835\udc61, 3\ud835\udc61\ud835\udc61) 0 \u2264 \ud835\udc61\ud835\udc61 \u2264 3. 
a) 8 
b) 81 
c) 162 
d) 324 
e) Nenhuma das anteriores. 
 
Questão 6 (1,0 ponto) 
Determine a equação do plano tangente à superfície S de equação 3\ud835\udc66\ud835\udc66\ud835\udc67\ud835\udc67 \u2212 \ud835\udc65\ud835\udc652\ud835\udc66\ud835\udc66 \u2212 \ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc672 = 3 no ponto (1,2,1). 
a) \u22125\ud835\udc65\ud835\udc65 \u2212 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 \u2212 3 = 0 
b) 5\ud835\udc65\ud835\udc65 \u2212 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 + 3 = 0 
c) 5\ud835\udc65\ud835\udc65 + 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 + 3 = 0 
d) \u22125\ud835\udc65\ud835\udc65 + 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 + 3 = 0 
e) Nenhuma das anteriores. 
 
Questão 7 (1,0 ponto) 
Calcule \u222c (\ud835\udc66\ud835\udc662 \u2212 3\ud835\udc65\ud835\udc65)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \u2227 \ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 + (8\ud835\udc66\ud835\udc66 \u2212 \ud835\udc60\ud835\udc607\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 \u2227 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + 5\ud835\udc67\ud835\udc67\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 \u2227 \ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udc46\ud835\udc46 através da superfície S que é o bordo do 
elipsoide \ud835\udc51\ud835\udc51
2
9
+ \ud835\udc66\ud835\udc662 + \ud835\udc67\ud835\udc672
25
\u2264 1 orientada com a normal que aponta para dentro. 
a) 60\ud835\udf0b\ud835\udf0b 
b) \u2212200\ud835\udf0b\ud835\udf0b 
c) \u2212240\ud835\udf0b\ud835\udf0b 
d) \u221236\ud835\udf0b\ud835\udf0b 
e) Nenhuma das anteriores. 
 
Questão 8 (1,0 ponto) 
Calcule a massa da placa plana descrita por \ud835\udc65\ud835\udc652 + \ud835\udc66\ud835\udc662 \u2264 9, \ud835\udc66\ud835\udc66 \u2265 0, com densidade \ud835\udeff\ud835\udeff = \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ufffd\ud835\udc65\ud835\udc652 + \ud835\udc66\ud835\udc662. 
a) \ud835\udf0b\ud835\udf0b 
b) 3\ud835\udf0b\ud835\udf0b 
c) 9\ud835\udf0b\ud835\udf0b 
d) 18\ud835\udf0b\ud835\udf0b 
e) Nenhuma das anteriores. 
 
Questão 9 (1,0 ponto) 
Calcule \u222b \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 \u2212 2\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udefe\ud835\udefe , sendo \ud835\udefe\ud835\udefe a curva 
esboçada ao lado, ligando o ponto (-3, 0) ao ponto (3, 0). 
a) 1 
b) 6 
c) \ud835\udf0b\ud835\udf0b 
d) -6 
e) Nenhuma das anteriores. 
 
 
 
Questão 10 (1,0 ponto) 
Sobre a função \ud835\udc53\ud835\udc53(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) = \u22128\ud835\udc65\ud835\udc652 \u2212 4\ud835\udc66\ud835\udc662 + \ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66 \u2212 7 é correto afirmar: 
a) (0,0) é seu único ponto crítico e é um ponto de máximo local. 
b) (0,0) é seu único ponto crítico e é um ponto de mínimo local. 
c) (0,0) é seu único ponto crítico e não é nem ponto de máximo nem de mínimo local. 
d) (1,1) é seu único ponto crítico e não é nem ponto de máximo nem de mínimo local. 
e) Nenhuma das anteriores. 
 
3 
 
GABARITO 
 
curso: Engenharia de Produção/ Engenharia de Computação bimestre: 4
o bimestre P3 
 
Disciplina: MCA502 - Cálculo II 
 
Questão 1 
Alternativa C: \ud835\udc51\ud835\udc51\ud835\udf15\ud835\udf15
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61) = \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61), \ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61)) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61) + \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(\ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61), \ud835\udc66\ud835\udc66(\ud835\udc61\ud835\udc61)) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(\ud835\udc61\ud835\udc61)\u21d2 
 \ud835\udc51\ud835\udc51\ud835\udf15\ud835\udf15
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(5) = \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(2,4) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(5) + \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51
(2,4) \u2219 \ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51
(5) = 3 \u2219 (\u22121) + 2 \u2219 2 = 1 
Assim no instante t = 5 a temperatura no besouro está aumentando 1 unidades por segundo. 
 
Questão 2 
Alternativa A: Massa \u222c \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66)\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 =\ud835\udc45\ud835\udc45 \u222b \u222b \ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc653042 = \u222b 13 \ud835\udc66\ud835\udc663|3042 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 = \u222b 9\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 = 1842 
 
Questão 3 
Alternativa A: Massa = \u222b \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc50\ud835\udc50 = \u222b \u2016\u210e\u2032(\ud835\udc61\ud835\udc61)\u2016\ud835\udf0b\ud835\udf0b0\u210e \ud835\udeff\ud835\udeff\ufffd\u210e(\ud835\udc61\ud835\udc61)\ufffd\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 
\u210e(\ud835\udc61\ud835\udc61) = (3\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, 3\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, 4\ud835\udc61\ud835\udc61)) \u21d2\u210e\u2032(\ud835\udc61\ud835\udc61) = (\u22123\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc61\ud835\udc61, 3\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc61\ud835\udc61, 4) 
\u2016\u210e\u2032(\ud835\udc61\ud835\udc61)\u2016 = \ufffd9\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc602\ud835\udc61\ud835\udc61 + 9\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc502\ud835\udc61\ud835\udc61 + 16 = \u221a25 = 5 
e também \ud835\udeff\ud835\udeff\ufffd\u210e(\ud835\udc61\ud835\udc61)\ufffd = 4\ud835\udc67\ud835\udc67 = 16\ud835\udc61\ud835\udc61 
Massa= \u222b 5 \u2219\ud835\udf0b\ud835\udf0b20 16\ud835\udc61\ud835\udc61\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 = 40\ud835\udc61\ud835\udc612|\ud835\udf0b\ud835\udf0b20 = 10\ud835\udf0b\ud835\udf0b2 
 
Questão 4 
Alternativa B: \ud835\udefe\ud835\udefe(\ud835\udc61\ud835\udc61) = (\ud835\udc61\ud835\udc613, 5\ud835\udc61\ud835\udc612, \ud835\udc61\ud835\udc61)\u21d2 \ud835\udefe\ud835\udefe(2) = (8, 20, 2) 
\ud835\udefe\ud835\udefe\u2032(\ud835\udc61\ud835\udc61) = (3\ud835\udc61\ud835\udc612, 10\ud835\udc61\ud835\udc61, 1) 
e 
\ud835\udefe\ud835\udefe\u2032(2) = (12, 20, 1) 
A reta tangente é \ud835\udc4b\ud835\udc4b(\ud835\udc50\ud835\udc50) = (8, 20, 2) + \ud835\udc50\ud835\udc50(12, 20, 1) = (8 + 12\ud835\udc50\ud835\udc50, 20 + 20\ud835\udc50\ud835\udc50, 2 + \ud835\udc50\ud835\udc50), \ud835\udc50\ud835\udc50 \u2208 \u211d 
 
Questão 5 
Alternativa D: Sendo \ud835\udefe\ud835\udefe(\ud835\udc61\ud835\udc61) = (2\ud835\udc61\ud835\udc61, 3\ud835\udc61\ud835\udc61, 3\ud835\udc61\ud835\udc61) 0 \u2264 \ud835\udc61\ud835\udc61 \u2264 3 
temos \ud835\udefe\ud835\udefe\u2032(\ud835\udc61\ud835\udc61) = (2, 3, 3) 
\ud835\udf0f\ud835\udf0f = \ufffd 2\ud835\udc66\ud835\udc66\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + \ud835\udc65\ud835\udc652\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 + \ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc67\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67
\ud835\udefe\ud835\udefe
= \ufffd (6\ud835\udc61\ud835\udc61 \u2219 2 + 4\ud835\udc61\ud835\udc612 \u2219 3 + 2\ud835\udc61\ud835\udc61 \u2219 3\ud835\udc61\ud835\udc61 \u2219 3)\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 = \ufffd (12\ud835\udc61\ud835\udc61 + 30\ud835\udc61\ud835\udc612)\ud835\udc51\ud835\udc51\ud835\udc61\ud835\udc61 = (6\ud835\udc61\ud835\udc612 + 10\ud835\udc61\ud835\udc613)|30 =3030 = 54 + 270 = 324 
 
Questão 6 
Alternativa E: A superfície S é a superfície de nível 3 da função \ud835\udc53\ud835\udc53(\ud835\udc65\ud835\udc65) = 3\ud835\udc66\ud835\udc66\ud835\udc67\ud835\udc67 \u2212 \ud835\udc65\ud835\udc652\ud835\udc66\ud835\udc66 \u2212 \ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc672. 
O gradiente de \ud835\udc53\ud835\udc53 num ponto genérico é dado por \u2207\ud835\udc53\ud835\udc53\ufffd\ufffd\ufffd\ufffd\u20d7 (\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66, \ud835\udc67\ud835\udc67) = (\u22122\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66 \u2212 \ud835\udc67\ud835\udc672,\u2212 \ud835\udc65\ud835\udc652 + 3\ud835\udc67\ud835\udc67, 3\ud835\udc66\ud835\udc66 \u2212 2\ud835\udc65\ud835\udc65\ud835\udc67\ud835\udc67). 
No ponto (1,2,1) obtemos \u2207\ud835\udc53\ud835\udc53\ufffd\ufffd\ufffd\ufffd\u20d7 (1,2,1) = (\u22125, 2, 4), que é perpendicular à S e ao seu plano tangente no ponto. 
Assim o plano terá equação \u22125\ud835\udc65\ud835\udc65 + 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 + \ud835\udc51\ud835\udc51 = 0. 
Como (1,2,1) é um ponto do plano \u22125 + 4 + 4 + \ud835\udc51\ud835\udc51 = 0 \u21d2 \ud835\udc51\ud835\udc51 = \u22123, logo o plano é \u22125\ud835\udc65\ud835\udc65 + 2\ud835\udc66\ud835\udc66 + 4\ud835\udc67\ud835\udc67 \u2212 3 = 0. 
 
 
4 
 
Questão 7 
Alternativa B: 
Pelo Teorema de Gauss vale: 
\ud835\udc3c\ud835\udc3c = \u222c (\ud835\udc66\ud835\udc662 \u2212 3\ud835\udc65\ud835\udc65)\ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \u2227 \ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 + (8\ud835\udc66\ud835\udc66 \u2212 \ud835\udc60\ud835\udc607\ud835\udc67\ud835\udc67)\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67 \u2227 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 + 5\ud835\udc67\ud835\udc67\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65 \u2227 \ud835\udc51\ud835\udc51\ud835\udc66\ud835\udc66 \ud835\udc46\ud835\udc46 = \u2212 \u222d (\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ud835\udc37\ufffd\u20d7\ufffd\ud835\udc39\ud835\udc49\ud835\udc49 )\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 == \u2212\u222d (\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51 + \ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udc51\ud835\udc51 + \ud835\udf15\ud835\udf15\ud835\udc45\ud835\udc45\ud835\udf15\ud835\udf15\ud835\udc67\ud835\udc67\ud835\udc49\ud835\udc49 )\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 
O sinal de menos na frente da integral tripla é devido à orientação \u201cpara dentro\u201d. 
\ud835\udc3c\ud835\udc3c = \u2212\ufffd (\u22123 + 8 + 5)\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51 = \u221210\ud835\udc51\ud835\udc51\ud835\udc50\ud835\udc50\ud835\udc49\ud835\udc49(\ud835\udc51\ud835\udc51) = \u221210 \u2219 43 .\ud835\udf0b\ud835\udf0b. \ud835\udc4e\ud835\udc4e \u2219 \ud835\udc4f\ud835\udc4f \u2219 \ud835\udc50\ud835\udc50 = \u221210 \u2219 43 .\ud835\udf0b\ud835\udf0b. 3 \u2219 1 \u2219 5 = \u2212200\ud835\udf0b\ud835\udf0b\ud835\udc49\ud835\udc49 
 
Questão 8 
Alternativa C: 
Massa = \u222c \ud835\udeff\ud835\udeff(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66)\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc37\ud835\udc37 = \u222c \ufffd\ud835\udc65\ud835\udc652 + \ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc37\ud835\udc37 
Vamos usar coordenadas polares no cálculo da integral dupla 
Parametrização: 
\ud835\udc65\ud835\udc65 = \ud835\udc5f\ud835\udc5f\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc5f\ud835\udc5f 0 \u2264 \ud835\udc5f\ud835\udc5f \u2264 \ud835\udf0b\ud835\udf0b , (\ud835\udc66\ud835\udc66 \u2265 0) 
\ud835\udc66\ud835\udc66 = \ud835\udc5f\ud835\udc5f\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc5f\ud835\udc5f 0 \u2264 \ud835\udc5f\ud835\udc5f \u2264 3 
Jacobiano = \ud835\udc5f\ud835\udc5f 
\u222c \ufffd\ud835\udc65\ud835\udc652 + \ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc51\ud835\udc37\ud835\udc37 = \u222b \u222b \ud835\udc5f\ud835\udc5f \u2219 \ud835\udc5f\ud835\udc5f\ud835\udc51\ud835\udc51\ud835\udc67\ud835\udc67\ud835\udc51\ud835\udc51\ud835\udc5f\ud835\udc5f = \u222b 13 \ud835\udc5f\ud835\udc5f3|30\ud835\udc51\ud835\udc51\ud835\udc5f\ud835\udc5f =\ud835\udf0b\ud835\udf0b030\ud835\udf0b\ud835\udf0b0 = \u222b 9\ud835\udf0b\ud835\udf0b0 \ud835\udc51\ud835\udc51\ud835\udc5f\ud835\udc5f = 9\ud835\udf0b\ud835\udf0b. 
 
Questão 9 
Alternativa B: 
Como a equação da curva não é fornecida, não conseguimos aplicar a definição de integral de linha. 
O campo é conservativo. Vamos obter sua função potencial: 
\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc65\ud835\udc65
(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662 \u21d2 \ud835\udf15\ud835\udf15(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662 + \ud835\udc54\ud835\udc54(\ud835\udc66\ud835\udc66) 
\ud835\udf15\ud835\udf15\ud835\udf15\ud835\udf15
\ud835\udf15\ud835\udf15\ud835\udc66\ud835\udc66
(\ud835\udc65\ud835\udc65, \ud835\udc66\ud835\udc66) = \u22122\ud835\udc65\ud835\udc65\ud835\udc66\ud835\udc66\ud835\udc50\ud835\udc50\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc60\ud835\udc66\ud835\udc662 \u21d2 \ud835\udf15\ud835\udf15(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662 + \u210e(\ud835\udc65\ud835\udc65) 
Comparando as duas expressões, concluímos que \ud835\udf15\ud835\udf15(\ud835\udc65\ud835\udc65,\ud835\udc66\ud835\udc66) = \ud835\udc65\ud835\udc65\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662 é uma função potencial 
\ufffd \ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc50\ud835\udc66\ud835\udc662\ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65