Question

Determine a massa de uma lâmina, dada pela integral, onde (2 + x + y) representa a variação da densidade.
Escolha uma:
a. 3
b. 23/3
c. 7/3
d. 11/3
e. 7

Answer 1

Boa tarde Aline !

Solução!

$Temos~~~~D: \begin{cases} 0 \leq x \leq 1\\\ 0 \leq y \leq \leq 2-2x \end{cases}\\\\\\\\\\\\ M=\displaystyle\int\int\delta(x,y)dA=\displaystyle \int_{0} ^{1} \int_{0}^{2-2x}(2+x+y)dydx\\\\\\\\\\ integrando~~em~~y~~fica ~~assim\\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[2y+2xy+ \frac{y^{2} }{2}\bigg] ^{2-2x}_{0}\\\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[2(2-2x)+2x(2-2x)+ \frac{(2-2x)^{2} }{2}\bigg]dx \\\\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[(2-2x).\bigg(2+x+ \frac{2-2x}{2}\bigg) \bigg]dx$

$\displaystyle \int_{0} ^{1}\bigg[(2-2x).\bigg( \frac{4+2x+2-2x}{2}\bigg) \bigg]dx\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[(2-2x).\bigg( \frac{6}{2}\bigg) \bigg]dx\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[(2-2x).(3)\bigg] dx\\\\\\\ \displaystyle \int_{0} ^{1}\bigg[(6-6x)\bigg] dx\\\\\\\ \bigg[6x- \dfrac{6 x^{2} }{2} )\bigg]_{0}^{1}\\\\\\\ \bigg[6x- 3x^{2} \bigg]_{0}^{1}\\\\\\\ \bigg[6(1)- 3(1)^{2} \bigg]_{0}^{1}=\bigg[6-3\bigg]=3$


$\boxed{Resposta: ~~Massa=3~~\boxed{Alternativa~~A}}$

Boa tarde!
Bons estudos!

Answer 2

Resposta:

Boa Tarde,

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