Question

Calcule a integral tripla com fundo preto:

Answer 1

Boa noite Oliveira!

Solução!

$\displaystyle\int\displaystyle\int\int12xy^{2}z^{3}dv\\\~~~~ _{G}$

$Intervalos!\\\\\\ -1 \leq x \leq 2~~,0 \leq y \leq 3~~,0 \leq z \leq 2$


$\displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3} \int_{0}^{2} 12xy^{2}z^{3}dzdydx\\\\\\\ \displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3}12xy^{2} \bigg[z^{3}\bigg]_{0} ^{2}dydx\\\\\\\\ \displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3}12xy^{2} \bigg[ \frac{z ^{4} }{4} \bigg]_{0} ^{2}dydx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3}12xy^{2} \bigg[ ( 2)^{4} \bigg]dydx$


$\dfrac{1}{4}\displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3}12xy^{2} \bigg[ 16 \bigg]dydx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3}192xy^{2} dydx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2}192x\bigg[y^{2}\bigg]_{0}^{3} dx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2}192x\bigg[ \frac{y^{3} }{3} \bigg]_{0}^{3} dx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2}192x\bigg[ \frac{(3)^{3} }{3} \bigg] dx$


$\dfrac{1}{4}\displaystyle\int_{-1} ^{2}192x\bigg[ 9\bigg] dx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2}1728x dx\\\\\\\\ \dfrac{1}{4}\displaystyle\int_{-1} ^{2}192x\bigg[ 9\bigg] dx\\\\\\\\ \dfrac{1}{4}.1728\bigg[x\bigg]_{-1}^{2}\\\\\\\\ \dfrac{1}{4}.1728\bigg[ \frac{ x^{2} }{2} \bigg]_{-1}^{2}\\\\\\\\ \dfrac{1}{4}.1728. \dfrac{1}{2} \bigg[ (2)^{2}- (-1)^{2} \bigg]\\\\\\\\ \dfrac{1}{4}.1728. \dfrac{1}{2} \bigg[ 4- 1 \bigg]$


$\dfrac{1}{4}.1728. \dfrac{1}{2} \bigg[ 3 \bigg]\\\\\\\\ \dfrac{1\times1728\times3}{8}\\\\\\ \dfrac{5184}{8}=648$



$\boxed{Resposta:~~\displaystyle\int_{-1} ^{2} \displaystyle\int_{0}^{3} \int_{0}^{2} 12xy^{2}z^{3}dzdydx=648}$

Boa noite!

Bons estudos!