Question

URGENTE, AJUDA

Com base no calculo de integrais, calcule a integral abaixo e, posteriormente assinale a alternativa correta.

∫∫rF(X,Y) dxdy, sendo f(x,y)= x²+y² definida retangular R=[0,1] X [0,1];

Answer 1

$R=\left\{(x,y)\in\mathbb{R}^2|0\leq x\leq 1, 0\leq y\leq 1\right\}$
Sabemos que para esse tipo de região podemos fazer
$\boxed{\iint_{R} f(x,y)dxdy=\iint_{R}f(x,y)dydx}$
logo:
$\displaystyle i)~~~~\iint_{R}f(x,y)dxdy=\iint\limits_{0~0}^{~~~~1~1}\left(x^2+y^2\right)dxdy\\\\ii)~~~\iint\limits_{0~0}^{~~~~1~1}\left(x^2+y^2\right)dxdy=\int\limits_{0}^{1}\left[\int\limits_{0}^{1}\left(x^2+y^2\right)dx\right]dy\\\\iii)~~\int\limits_{0}^{1}\left[\int\limits_{0}^{1}\left(x^2+y^2\right)dx\right]dy=\int\limits_{0}^{1}\left[\frac{x^3}{3}+y^2x\right]_{0}^{1}dy=\int\limits_{0}^{1}\left(\frac{1}{3}+y^2\right)dy\\\\ iv)~~\int\limits_{0}^{1}\left(\frac{1}{3}+y^2\right)dy=\left[\frac{y}{3}+\frac{y^3}{3}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{3}=\boxed{\frac{2}{3}}$

Caso haja alguma duvida comentar abaixo. Bons estudos
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