Question

O volume do sólido obtido girando-se em relação ao eixo x, a região limitada pelas retas y=0, x=1, x=4 e y = x - 1 é:

a) 10 π
b) 7 π
c) 11 π
d) 9 π
e) 8 π

Answer 1

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Solução!


$\boxed{V= \pi \displaystyle \int _{a}^{b} [f(x) ]^{2} dx}$


$V= \pi \displaystyle \int _{1}^{4} [f(x-1) ]^{2} dx\\\\\\\ V= \pi \displaystyle \int _{1}^{4} ( x^{2} -2x+1) dx\\\\\\\ V= \pi\bigg( \frac{ x^{3} }{3} - x^{2} +x\bigg)\bigg|_{1}^{4}\\\\\\\ V= \pi\bigg( \dfrac{ x^{3} }{3} - x^{2} +x\bigg)-\bigg( \dfrac{ x^{3} }{3} - x^{2} +x\bigg)\bigg|_{1}^{4}\\\\\\\\\\\ V= \pi\bigg( \dfrac{ (4)^{3} }{3} - (4)^{2} +(4)\bigg)-\bigg ( \dfrac{ (1)^{3} }{3} - (1)^{2} +(1)\bigg)$



$V= \pi\bigg( \dfrac{64}{3} - 16 +4\bigg)-\bigg ( \dfrac{1 }{3} - 1 +1\bigg)\\\\\\\\\ V= \pi\bigg( \dfrac{64-48+12}{3}\bigg)-\bigg ( \dfrac{1+3-3 }{3} \bigg)\\\\\\\\\ V= \pi\bigg( \dfrac{28}{3}\bigg)-\bigg ( \dfrac{1 }{3} \bigg)\\\\\\\\\ V= \pi\bigg( \dfrac{28}{3}-\dfrac{1 }{3} \bigg)$



$V= \pi\bigg( \dfrac{27}{3} \bigg)\\\\\\\ V=9 \pi$



$\boxed{Resposta:~~V=9 \pi ~~\boxed{Alternativa~~D}}$


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Bons estudos!

Answer 2

Resposta:

Explicação passo-a-passo: