Question

As integrais duplas podem ser aplicadas no cálculo da área de figuras planas e do volume de sólidos, desde que considerando funções e regiões convenientes. Qual o volume do sólido limitado superiormente pela superfície representada pela função f(x,y) = xy e inferiormente pela região retangular R = [0,1]x[0,1] no plano xy? Alternativas: a) 1/4. b) 1/2. c) 1. d) 3/2. e) 2.

Answer 1

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Solução!

$V=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}dydx\\\\\\\\ V=\displaystyle\int_{0}^{1}x\bigg[y\bigg]_{0}^{1}dx\\\\\\\\ V=\displaystyle\int_{0}^{1}x\bigg[ \frac{ y^{2} }{2} \bigg]_{0}^{1}dx\\\\\\\\ V= \dfrac{1}{2} \displaystyle\int_{0}^{1}x\bigg[y^{2} \bigg]_{0}^{1}dx\\\\\\\\ V= \dfrac{1}{2} \displaystyle\int_{0}^{1}x\bigg[y^{2} - y^{2} \bigg]_{0}^{1}dx\\\\\\\\ V= \dfrac{1}{2} \displaystyle\int_{0}^{1}x\bigg[1^{2} - 0^{2} \bigg]_{0}^{1}dx$

$V= \dfrac{1}{2} \displaystyle\int_{0}^{1}x\bigg[1 \bigg]dx\\\\\\\\ V= \dfrac{1}{2} \displaystyle\int_{0}^{1}xdx\\\\\\\\ V= \dfrac{1}{2}\bigg[x \bigg]_{0}^{1}\\\\\\\\ V= \dfrac{1}{2}\bigg[ \frac{ x^{2} }{2} \bigg]_{0}^{1}\\\\\\\\ V= \dfrac{1}{2}. \dfrac{1}{2} \bigg[ x^{2}- x^{2} \bigg]_{0}^{1}\\\\\\\\ V= \dfrac{1}{2}. \dfrac{1}{2} \bigg[ 1- 0\bigg]\\\\\\\ V= \dfrac{1}{2}. \dfrac{1}{2} \bigg[ 1\bigg]\\\\\\\ V= \dfrac{1}{2}. \dfrac{1}{2}\\\\\\ V= \dfrac{1}{4}u~~v$

$\boxed{Resposta:~~V=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}dydx= \dfrac{1}{4}u.v ~~~\boxed{Alternativa~~A}}$

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