Ensino Superior ⇒ Integral tripla Tópico resolvido

[magben]

[tex3]\int\limits_{0}^{4}\int\limits_{0}^{8-2x}\int\limits_{0}^{4-\frac{y}{2}-x}xdxdydz[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{0}^{4}\int\limits_{0}^{8-2x}\int\limits_{0}^{4-\frac{y}{2}-x}xdxdydz[/tex3]

Resposta

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[tex3]\frac{64}{3}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{64}{3}[/tex3]

[παθμ]

magben,

Faça primeiro a integral em z. Fica:

[tex3]I=\int_{0}^{4}\int_{0}^{8-2x}x\left(4-\frac{y}{2}-x\right)dx \; dy=\int_{0}^ {4}\int_{0}^{8-2x}\left(4x-x^2-\frac{xy}{2}\right)dx \; dy.[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]I=\int_{0}^{4}\int_{0}^{8-2x}x\left(4-\frac{y}{2}-x\right)dx \; dy=\int_{0}^ {4}\int_{0}^{8-2x}\left(4x-x^2-\frac{xy}{2}\right)dx \; dy.[/tex3]

Agora faça a integral em y:

[tex3]I=\int_{0}^{4}\left(4x(8-2x)-x^2(8-2x)-\frac{x}{2} \cdot \frac{(8-2x)^2}{2}\right)dx=\int_{0}^{4}(x^3-8x^2+16x)dx=64-\frac{8 \cdot 64}{3}+16 \cdot \frac{16}{2}=\boxed{\frac{64}{3}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]I=\int_{0}^{4}\left(4x(8-2x)-x^2(8-2x)-\frac{x}{2} \cdot \frac{(8-2x)^2}{2}\right)dx=\int_{0}^{4}(x^3-8x^2+16x)dx=64-\frac{8 \cdot 64}{3}+16 \cdot \frac{16}{2}=\boxed{\frac{64}{3}}[/tex3]