Question

integral tripla. calculo 2.

Answer 1

Resposta:

a) 8,667

b) 256

Explicação passo-a-passo:

Para resolver essas integrais precisamos saber algumas propriedades:

Integrar no volume "dv" é pode ser substituido pelo volume em "dx.dy.dz"

A integral qualquer "du" de uma contante (um número qualquer) é "u".

a)

$\int\limits^a_b {\int\limits^c_d {\int\limits^e_f {x.y.z^{2}} \, dx } \, dy} \, dz$

$\int\limits^3_1 {\int\limits^2_0 {\int\limits^1_0 {x.y.z^{2}} \, dx } \, dy} \, dz\\\int\limits^3_1 {\int\limits^2_0 { {\frac{1}{2} x^{2}.y.z^{2}}} \, dy} \, dz\\\frac{1}{2} .[1^{2}-0^{2}]\int\limits^3_1 {\int\limits^2_0 { {y.z^{2}}} \, dy} \, dz\\\frac{1}{2} .1\int\limits^3_1 {\frac{1}{2}.y^{2}.z^{2}} \, dz\\\frac{1}{4} .[2^{2}-0^{2}]\int\limits^3_1 {z^{2}} \, dz\\\frac{1}{4} .4\int\limits^3_1 {z^{2}} \, dz\\ \frac{1}{3}.z^{3}}$

$\frac{1}{3} [3^{3}-1^{3}]\\\\\frac{1}{3} [27-1]\\\\\frac{1}{3} 26\\\\\frac{26}{3}$

8,667

b)

$\int\limits^a_b {\int\limits^c_d {\int\limits^e_f {x} \, dx } \, dy } \, dz$

$\int\limits^4_0 {\int\limits^8_0 {\int\limits^4_0 {x} \, dx } \, dy } \, dz\\\int\limits^4_0 {\int\limits^8_0 {\frac{1}{2} x^{2}} \, dy } \, dz\\\frac{1}{2} [4^{2}-0^{2}]\int\limits^4_0 {\int\limits^8_0 {1} \, dy } \, dz\\\frac{16}{2} \int\limits^4_0 {y } \, dz\\8.[8-0] \int\limits^4_0 {1} \, dz\\64 z\\\\64.[4-0]\\\\256$