Question

Usando a técnica da integral dupla, encontre o volume do sóli do gerado pela expressão ∫ ∫(x2 + y2) dxdy para os intervalos R=[- 1,1] x[-2,1

Answer 1

Vamos calcular a integral resolvendo a questão: 



$V= \int\limits^{-1}_1 { \int\limits^{-2}_1 {(x^{2}+y^{2})} \, dx } \, dy \\\\\\\ V= \int\limits^{-1}_1 {dx.[ \int\limits^{-2}_1 {(x^{2}+y^{2})} \, dy } ]\\\\\\V= \int\limits^{-1}_1 {} \, dx .[ x^2.y+ \frac{y^3}{3} ]^{-2}_1\\\\\\V= \int\limits^{-1}_1 {dx} \, .[(x^{2}.-2+ \frac{(-2)^{3}}{3} )-(x^{2}.1+ \frac{(1)^{3}}{3} )]\\\\\\ V=\int\limits^{-1}_1 {dx} .[(-2x^{2}- \frac{8}{3} )-(x^{2}+ \frac{1}{3} )]\\\\\\V= \int\limits^{-1}_1 {dx} .[-2x^{2}-x^{2}- \frac{8}{3} - \frac{1}{3} ]$
$V= \int\limits^{-1}_1 {dx}.[-3x^{2}- \frac{9}{3} ]\\\\\\V= \int\limits^{-1}_1 {dx} .[-3x^{2}-3]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (resolvendo\ a\ segunda\ parte)\\\\\\V= \int\limits^{-1}_1 {(-3x^{2}-3)} \, dx \\\\\\V=[ \frac{-3x^{3}}{3} -3x ]^{-1}_1 \\\\\\V=[-x^{3}-3x]^{-1}_1 \\\\\\V=[-(-1)^{3}-3.(-1)]-[-(1)^{3}-3.(1)]\\\\\\V=[1+3]-[-1-3]\\\\\\V=[4]-[-4]\\\\\\V=4+4\\\\\\\boxed{\boxed{V=8\ u.v}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Resposta\ :\ 8\ u.v\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$