Calcule o volume do conjunto dado. (Sugerimos ao leitor desenhar o conjunto.)

g) (x^2 / a^2) + (y^2 / b^2) + (z^2 / c^2) ≤ 1 . ( a > 0, b > 0 e c > 0 ).

Passo 1

Nesse problema iremos calcular o volume do conjunto através da integral tripla. Para isso, vamos reduzir o problema de uma integral tripla para dupla, encontrando o novo conjunto K e os intervalos de integração. Vamos lá!

Passo 2

Temos que o volume é dado pela integral

∫ ∫ ∫ B d x ⋅ d y ⋅ d z = ∫ ∫ K ∫ a b d z ⋅ d x ⋅ d y

Substituindo nossos intervalos, chegamos que

∫ ∫ K ∫ - c ⋅ 1 - x 2 a 2 - y 2 b 2 c ⋅ 1 - x 2 a 2 - y 2 b 2 d z ⋅ d x ⋅ d y = 2 ⋅ ∫ ∫ K ∫ 0 c ⋅ 1 - x 2 a 2 - y 2 b 2 d z ⋅ d x ⋅ d y =

= 2 ⋅ c ⋅ ∫ ∫ K 1 - x 2 a 2 - y 2 b 2 ⋅ d x ⋅ d y

Com nosso K sendo x 2 a 2 + y 2 b 2 ≤ 1 . Assim, aplicando uma transformação para coordenadas polares, temos que

x = a ⋅ ρ ⋅ cos ⁡ θ

y = b ⋅ ρ ⋅ sin ⁡ θ

0 ≤ ρ ≤ 1

0 ≤ θ ≤ 2 ⋅ π

Com o jacobiano sendo d x ⋅ d y = a ⋅ b ⋅ ρ ⋅ d ρ ⋅ d θ . Substituindo na integral e resolvendo, chegamos que

2 ⋅ c ⋅ ∫ 0 2 ⋅ π ∫ 0 1 1 - ρ 2 ⋅ a ⋅ b ⋅ ρ ⋅ d ρ ⋅ d θ = 2 ⋅ a ⋅ b ⋅ c ⋅ ∫ 0 2 ⋅ π d θ ⋅ ∫ 0 1 1 - ρ 2 ⋅ ρ ⋅ d ρ =

Aplicando uma mudança de variável, chamando u = 1 - ρ 2 , temos que

= 4 ⋅ π ⋅ a ⋅ b ⋅ c ⋅ ∫ 1 0 - 1 2 ⋅ u 1 2 ⋅ d u = - 2 ⋅ π ⋅ a ⋅ b ⋅ c ⋅ 2 3 ⋅ u 3 2 1 0 =

= - 4 ⋅ π ⋅ a ⋅ b ⋅ c 3 ⋅ 0 3 2 - 1 3 2 = 4 ⋅ π ⋅ a ⋅ b ⋅ c 3

Resposta

∫ ∫ ∫ B d x ⋅ d y ⋅ d z = 4 ⋅ π ⋅ a ⋅ b ⋅ c 3

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